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GRADE 10 PRACTICE

Physics Problems: Direct current, Grade 10

Laws of circuits, current operation and connections.

23 problems across 4 sub-topics
Practise in the interactive mode

Here the conditions and analysis of the solution can be read without an account. In the simulator, you need to collect the same analyzes yourself: “Given,” formulas, order of steps and answer.

Problem statements

The answer options are there to check against, and the walkthrough opens under each problem.

23 problems
Position in this set: 1 from 23

Problem #106

Direct current: Ohm's law and resistance

To a steel conductor length 80\,\text{см} and cross section 0{,}15\,\text{мм}^{2} on 1\,\text{мин} applied voltage 1{,}2\,\text{В}. Resistivity 0{,}12\,\text{Ом}\cdot\text{мм}^{2}/\text{м}. Define resistance, current, power, work, heat and charge.

Answer options

  1. \left(0{,}64\,\text{Ом};\ 1{,}875\,\text{А};\ 2{,}25\,\text{Вт};\ 135\,\text{Дж};\ 135\,\text{Дж};\ 112{,}5\,\text{Кл}\right)
  2. \left(0{,}32\,\text{Ом};\ 0{,}9375\,\text{А};\ 1{,}125\,\text{Вт};\ 67{,}5\,\text{Дж};\ 67{,}5\,\text{Дж};\ 56{,}25\,\text{Кл}\right)
  3. \left(1{,}28\,\text{Ом};\ 3{,}75\,\text{А};\ 4{,}5\,\text{Вт};\ 270\,\text{Дж};\ 270\,\text{Дж};\ 225\,\text{Кл}\right)
  4. \left(6{,}4\,\text{Ом};\ 18{,}75\,\text{А};\ 22{,}5\,\text{Вт};\ 1350\,\text{Дж};\ 1350\,\text{Дж};\ 1125\,\text{Кл}\right)
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Given
\ell
80\,\text{см}0{,}8\,\text{м}
S
0{,}15\,\text{мм}^{2}
t
1\,\text{мин}60\,\text{с}
U
1{,}2\,\text{В}
\rho
0{,}12\,\tfrac{\text{Ом}\cdot\text{мм}^{2}}{\text{м}}
Find

RIPAQq

  1. R = \dfrac{\rho\ell}{S}R = \dfrac{0{,}12\cdot 0{,}8}{0{,}15}R = 0{,}64\,\text{Ом}
  2. I = \dfrac{U}{R}I = \dfrac{1{,}2}{0{,}64}I = 1{,}875\,\text{А}
  3. P = IUP = 1{,}875\cdot 1{,}2P = 2{,}25\,\text{Вт}
  4. A = Pt = QA = 2{,}25\cdot 60A = 135\,\text{Дж}

    All work done by the current on the resistor turns into heat.

  5. q = Itq = 1{,}875\cdot 60q = 112{,}5\,\text{Кл}

AnswerR = 0{,}64\,\text{Ом}I = 1{,}875\,\text{А}P = 2{,}25\,\text{Вт}A = 135\,\text{Дж}Q = 135\,\text{Дж}q = 112{,}5\,\text{Кл}

Position in this set: 2 from 23

Problem #107

Direct Current: Circuits, Operation and Power

To the source with EMF 4{,}5\,\text{В} and internal resistance 2\,\text{Ом} connect the resistor for a minute 7\,\text{Ом}. Determine load current, voltage, power, work, heat, charge and number of electrons.

Answer options

  1. \left(5\,\text{А};\ 35\,\text{В};\ 17{,}5\,\text{Вт};\ 1050\,\text{Дж};\ 1050\,\text{Дж};\ 300\,\text{Кл};\ 1{,}875 \cdot 10^{21}\,\text{электронов}\right)
  2. \left(0{,}5\,\text{А};\ 3{,}5\,\text{В};\ 1{,}75\,\text{Вт};\ 105\,\text{Дж};\ 105\,\text{Дж};\ 30\,\text{Кл};\ 1{,}875 \cdot 10^{20}\,\text{электронов}\right)
  3. \left(0{,}25\,\text{А};\ 1{,}75\,\text{В};\ 0{,}875\,\text{Вт};\ 52{,}5\,\text{Дж};\ 52{,}5\,\text{Дж};\ 15\,\text{Кл};\ 9{,}375 \cdot 10^{19}\,\text{электронов}\right)
  4. \left(1\,\text{А};\ 7\,\text{В};\ 3{,}5\,\text{Вт};\ 210\,\text{Дж};\ 210\,\text{Дж};\ 60\,\text{Кл};\ 3{,}75 \cdot 10^{20}\,\text{электронов}\right)
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Given
\varepsilon
4{,}5\,\text{В}
r
2\,\text{Ом}
R
7\,\text{Ом}
t
1\,\text{мин}60\,\text{с}
Find

IUPAQ_1qn

  1. I = \dfrac{\varepsilon}{R + r}I = \dfrac{4{,}5}{7 + 2}I = 0{,}5\,\text{А}

    Ohm's law for a complete circuit.

  2. U = IRU = 0{,}5\cdot 7U = 3{,}5\,\text{В}
  3. P = IUP = 0{,}5\cdot 3{,}5P = 1{,}75\,\text{Вт}
  4. A = Pt,\quad Q_1 = I^{2}Rt\left(A;\ Q_1\right) = 1{,}75\cdot 60\left(A;\ Q_1\right) = 105\,\text{Дж}
  5. Q_2 = I^{2}rtQ_2 = 0{,}5^{2}\cdot 2\cdot 60Q_2 = 30\,\text{Дж}

    These are losses due to the internal resistance of the source.

  6. q = Itq = 0{,}5\cdot 60q = 30\,\text{Кл}
  7. n = \dfrac{q}{e}n = \dfrac{30}{1{,}6\cdot 10^{-19}}n = 1{,}875\cdot 10^{20}

AnswerI = 0{,}5\,\text{А}U = 3{,}5\,\text{В}P = 1{,}75\,\text{Вт}A = 105\,\text{Дж}Q_1 = 105\,\text{Дж}q = 30\,\text{Кл}n = 1{,}875\cdot 10^{20}

Position in this set: 3 from 23

Problem #108

Direct current: Ohm's law and resistance

Resistance of aluminum conductor at 100\,^{\circ}\text{C} equals 28{,}4\,\text{Ом}. What will it be like 500\,^{\circ}\text{C}, if \alpha=0{,}0042\,\text{К}^{-1}?

Answer options

  1. 124\,\text{Ом}
  2. 620\,\text{Ом}
  3. 62\,\text{Ом}
  4. 31\,\text{Ом}
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Given
t_1
100\,^{\circ}\text{C}
R_1
28{,}4\,\text{Ом}
t_2
500\,^{\circ}\text{C}
\alpha
0{,}0042\,\tfrac{1}{\text{К}}
Find

R_2

  1. R_1 = R_0\left(1 + \alpha t_1\right)\ \Rightarrow\ R_0 = \dfrac{R_1}{1 + \alpha t_1}R_0 = \dfrac{28{,}4}{1 + 0{,}0042\cdot 100}R_0 = 20\,\text{Ом}

    First we find the resistance at zero degrees.

  2. R_2 = R_0\left(1 + \alpha t_2\right)20\left(1 + 0{,}0042\cdot 500\right)R_2 = 62\,\text{Ом}

AnswerR_2 = 62\,\text{Ом}

Position in this set: 4 from 23

Problem #109

Direct Current: Circuits, Operation and Power

Under tension 3{,}2\,\text{В} the current in the circuit is 0{,}5\,\text{А}, and when 2{,}6\,\text{В} - 1\,\text{А}. Determine the emf and internal resistance of the source.

Answer options

  1. \left(1{,}9\,\text{В};\ 0{,}6\,\text{Ом}\right)
  2. \left(7{,}6\,\text{В};\ 2{,}4\,\text{Ом}\right)
  3. \left(38\,\text{В};\ 12\,\text{Ом}\right)
  4. \left(3{,}8\,\text{В};\ 1{,}2\,\text{Ом}\right)
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Given
U_1
3{,}2\,\text{В}
I_1
0{,}5\,\text{А}
U_2
2{,}6\,\text{В}
I_2
1\,\text{А}
Find

\varepsilonr

  1. \varepsilon = U_1 + I_1 r = U_2 + I_2 r

    The source emf is the same in both modes.

  2. U_1 - U_2 = r\left(I_2 - I_1\right)\ \Rightarrow\ r = \dfrac{U_1 - U_2}{I_2 - I_1}r = \dfrac{3{,}2 - 2{,}6}{1 - 0{,}5}r = 1{,}2\,\text{Ом}
  3. \varepsilon = U_1 + I_1 r\varepsilon = 3{,}2 + 0{,}5\cdot 1{,}2\varepsilon = 3{,}8\,\text{В}

Answer\varepsilon = 3{,}8\,\text{В}r = 1{,}2\,\text{Ом}

Position in this set: 5 from 23

Problem #110

Direct Current: Circuits, Operation and Power

Conduction electron concentration in germanium at room temperature 3\cdot10^{19}\,\text{м}^{-3}. What proportion of the total number of atoms are conduction electrons? Density 5400\,\text{кг/м}^{3}, molar mass 0{,}073\,\text{кг/моль}.

Answer options

  1. 6{,}76 \cdot 10^{-10}\,\text{доля}
  2. 3{,}38 \cdot 10^{-10}\,\text{доля}
  3. 1{,}352 \cdot 10^{-9}\,\text{доля}
  4. 6{,}76 \cdot 10^{-9}\,\text{доля}
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Given
n
3\cdot 10^{19}\,\tfrac{1}{\text{м}^{3}}
\rho
5400\,\tfrac{\text{кг}}{\text{м}^{3}}
M
0{,}073\,\tfrac{\text{кг}}{\text{моль}}
N_A
6\cdot 10^{23}\,\tfrac{1}{\text{моль}}
Find

\dfrac{n}{N}

  1. N = \dfrac{\rho N_A}{M}N = \dfrac{5400\cdot 6\cdot 10^{23}}{0{,}073}N = 4{,}44\cdot 10^{28}\,\text{м}^{-3}

    Here N is the number density of germanium atoms: the number of atoms per unit volume. The ratio n/N of the two number densities is dimensionless.

  2. \dfrac{n}{N}\dfrac{n}{N} = \dfrac{3\cdot 10^{19}}{4{,}44\cdot 10^{28}}\dfrac{n}{N} = 6{,}76\cdot 10^{-10}

Answer\dfrac{n}{N} = 6{,}76\cdot 10^{-10}

Position in this set: 6 from 23

Problem #111

Direct Current: Circuits, Operation and Power

How much aluminum will be released per day under current 140\,\text{кА} and voltage 4\,\text{В}? What is the energy consumption? Electrochemical equivalent 0{,}093\cdot10^{-6}\,\text{кг/Кл}.

Answer options

  1. \left(11250\,\text{кг};\ 4{,}84 \cdot 10^{11}\,\text{Дж}\right)
  2. \left(1125\,\text{кг};\ 4{,}84 \cdot 10^{10}\,\text{Дж}\right)
  3. \left(562{,}5\,\text{кг};\ 2{,}42 \cdot 10^{10}\,\text{Дж}\right)
  4. \left(2250\,\text{кг};\ 9{,}68 \cdot 10^{10}\,\text{Дж}\right)
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Given
t
1\,\text{сут}86\,400\,\text{с}
I
140\,\text{кА}140\cdot 10^{3}\,\text{А}
U
4\,\text{В}
k
0{,}093\cdot 10^{-6}\,\tfrac{\text{кг}}{\text{Кл}}
Find

mA

  1. m = kItm = 0{,}093\cdot 10^{-6}\cdot 140\cdot 10^{3}\cdot 86\,400m = 1125\,\text{кг}

    Faraday's law of electrolysis.

  2. A = IUtA = 140\cdot 10^{3}\cdot 4\cdot 86\,400A = 4{,}84\cdot 10^{10}\,\text{Дж} = 13\,440\,\text{кВт}\cdot\text{ч}

Answerm = 1125\,\text{кг}A = 4{,}84\cdot 10^{10}\,\text{Дж}

Position in this set: 7 from 23

Problem #112

Direct Current: Circuits, Operation and Power

Voltmeter on 6\,\text{В} has internal resistance 600\,\text{Ом}. What additional resistance must be connected in series to expand the measurement limit to 30\,\text{В}?

Answer options

  1. 4800\,\text{Ом}
  2. 24000\,\text{Ом}
  3. 2400\,\text{Ом}
  4. 1200\,\text{Ом}
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Given
U_0
6\,\text{В}
R_0
600\,\text{Ом}
U
30\,\text{В}
Find

R_{\text{д}}

  1. I_0 = \dfrac{U_0}{R_0}I_0 = \dfrac{6}{600}I_0 = 0{,}01\,\text{А}

    The same current flows through the additional resistor as through the voltmeter.

  2. U_{\text{д}} = U - U_0U_{\text{д}} = 30 - 6U_{\text{д}} = 24\,\text{В}
  3. R_{\text{д}} = \dfrac{U_{\text{д}}}{I_0}R_{\text{д}} = \dfrac{24}{0{,}01}R_{\text{д}} = 2400\,\text{Ом}

    The additional resistance is connected in series with the voltmeter.

AnswerR_{\text{д}} = 2400\,\text{Ом}

Position in this set: 8 from 23

Problem #113

Direct Current: Circuits, Operation and Power

The ammeter is designed for 2\,\text{А} and has internal resistance 0{,}4\,\text{Ом}. What resistance shunt should be connected in parallel to expand the measurement limit to 10\,\text{А}?

Answer options

  1. 0{,}05\,\text{Ом}
  2. 0{,}2\,\text{Ом}
  3. 1\,\text{Ом}
  4. 0{,}1\,\text{Ом}
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Given
I_0
2\,\text{А}
R_0
0{,}4\,\text{Ом}
I
10\,\text{А}
Find

R_{\text{ш}}

  1. U_0 = I_0 R_0 = U_{\text{ш}}U_0 = 2\cdot 0{,}4U_0 = 0{,}8\,\text{В}

    The shunt is connected in parallel, so the voltage is common.

  2. I_{\text{ш}} = I - I_0I_{\text{ш}} = 10 - 2I_{\text{ш}} = 8\,\text{А}
  3. R_{\text{ш}} = \dfrac{U_{\text{ш}}}{I_{\text{ш}}}R_{\text{ш}} = \dfrac{0{,}8}{8}R_{\text{ш}} = 0{,}1\,\text{Ом}

AnswerR_{\text{ш}} = 0{,}1\,\text{Ом}

Position in this set: 9 from 23

Problem #114

Direct Current: Circuits, Operation and Power

The crane motor operates under voltage 380\,\text{В} and current 40\,\text{А}. How long will it take him to lift a mass of 2\,\text{т} to the height 5\,\text{м} at efficiency 80\%?

Answer options

  1. 8{,}22\,\text{с}
  2. 4{,}11\,\text{с}
  3. 16{,}44\,\text{с}
  4. 82{,}2\,\text{с}
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Given
U
380\,\text{В}
I
40\,\text{А}
m
2\,\text{т}2000\,\text{кг}
h
5\,\text{м}
\eta
0{,}8
g
10\,\tfrac{\text{м}}{\text{с}^{2}}
Find

t

  1. A_{\text{п}} = mgh,\quad A_{\text{з}} = IUt

    Useful work is lifting a load, spent work is current work.

  2. \eta = \dfrac{A_{\text{п}}}{A_{\text{з}}} = \dfrac{mgh}{IUt}\ \Rightarrow\ t = \dfrac{mgh}{IU\eta}t = \dfrac{2000\cdot 10\cdot 5}{40\cdot 380\cdot 0{,}8}t = 8{,}22\,\text{с}

Answert=8{,}22\,\text{с}

Position in this set: 10 from 23

Problem #115

Direct Current: Circuits, Operation and Power

A light bulb 30\,\text{Вт}, 40\,\text{В} needs to be connected to the network 220\,\text{В}. What additional resistance should be connected in series?

Answer options

  1. 2400\,\text{Ом}
  2. 240\,\text{Ом}
  3. 120\,\text{Ом}
  4. 480\,\text{Ом}
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Given
P
30\,\text{Вт}
U_1
40\,\text{В}
U_2
220\,\text{В}
Find

R_{\text{д}}

  1. I = \dfrac{P}{U_1} = I_{\text{д}}I = \dfrac{30}{40}I = 0{,}75\,\text{А}
  2. U_{\text{д}} = U_2 - U_1U_{\text{д}} = 220 - 40U_{\text{д}} = 180\,\text{В}
  3. R_{\text{д}} = \dfrac{U_{\text{д}}}{I_{\text{д}}}R_{\text{д}} = \dfrac{180}{0{,}75}R_{\text{д}} = 240\,\text{Ом}

AnswerR_{\text{д}} = 240\,\text{Ом}

Position in this set: 11 from 23

Problem #116

Direct Current: Circuits, Operation and Power

It is necessary to make a heater with a power 600\,\text{Вт} for voltage 220\,\text{В}. What is the length of nichrome wire cross section 0{,}3\,\text{мм}^{2} needed when \rho=1{,}1\,\text{Ом}\cdot\text{мм}^{2}/\text{м}?

Answer options

  1. 44\,\text{м}
  2. 220\,\text{м}
  3. 22\,\text{м}
  4. 11\,\text{м}
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Given
P
600\,\text{Вт}
U
220\,\text{В}
S
0{,}3\,\text{мм}^{2}
\rho
1{,}1\,\tfrac{\text{Ом}\cdot\text{мм}^{2}}{\text{м}}
Find

\ell

  1. P = \dfrac{U^{2}}{R}\ \Rightarrow\ R = \dfrac{U^{2}}{P}
  2. R = \dfrac{\rho\ell}{S}\ \Rightarrow\ \dfrac{U^{2}}{P} = \dfrac{\rho\ell}{S}\ \Rightarrow\ \ell = \dfrac{U^{2}S}{P\rho}\ell = \dfrac{220^{2}\cdot 0{,}3}{600\cdot 1{,}1}\ell = 22\,\text{м}

Answer\ell = 22\,\text{м}

Position in this set: 12 from 23

Problem #225

Current, resistance and Ohm's law

Resistor voltage 12 V, resistance 4 Ohm. Find the current.

Answer options

  1. 3\,\text{А}
  2. 6\,\text{А}
  3. 1{,}5\,\text{А}
  4. 9\,\text{А}
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Given
U
12\,\text{В}
R
4\,\text{Ом}
Find

I

  1. I=\frac{U}{R}I=\frac{12}{4}I=3\,\text{А}

    We apply Ohm's law to a section of the circuit.

Answer3\,\text{А}

Position in this set: 13 from 23

Problem #226

Current, resistance and Ohm's law

Resistor voltage 24 V, resistance 6 Ohm. Find the current.

Answer options

  1. 8\,\text{А}
  2. 4\,\text{А}
  3. 2\,\text{А}
  4. 12\,\text{А}
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Given
U
24\,\text{В}
R
6\,\text{Ом}
Find

I

  1. I=\frac{U}{R}I=\frac{24}{6}I=4\,\text{А}

    We apply Ohm's law to a section of the circuit.

Answer4\,\text{А}

Position in this set: 14 from 23

Problem #227

Current, resistance and Ohm's law

Under tension 6 The ammeter shows 200 mA. Find the resistance of the circuit section.

Answer options

  1. 60\,\text{Ом}
  2. 30\,\text{Ом}
  3. 15\,\text{Ом}
  4. 90\,\text{Ом}
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Given
U
6\,\text{В}
i
200\,\text{мА}
Find

R

  1. I=\frac{i}{1000}I=\frac{200}{1000}I=0{,}2\,\text{А}

    Before calculating, we convert milliamps to amperes.

  2. R=\frac{U}{I}R=\frac{6}{0{,}2}R=30\,\text{Ом}

    We express resistance from Ohm's law.

Answer30\,\text{Ом}

Position in this set: 15 from 23

Problem #228

Current, resistance and Ohm's law

Under tension 12 The ammeter shows 300 mA. Find the resistance of the circuit section.

Answer options

  1. 80\,\text{Ом}
  2. 20\,\text{Ом}
  3. 40\,\text{Ом}
  4. 120\,\text{Ом}
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Given
U
12\,\text{В}
i
300\,\text{мА}
Find

R

  1. I=\frac{i}{1000}I=\frac{300}{1000}I=0{,}3\,\text{А}

    Before calculating, we convert milliamps to amperes.

  2. R=\frac{U}{I}R=\frac{12}{0{,}3}R=40\,\text{Ом}

    We express resistance from Ohm's law.

Answer40\,\text{Ом}

Position in this set: 16 from 23

Problem #229

Current, resistance and Ohm's law

Under tension 12 V through a resistor for 10 s charge passed 20 Cl. What current will be at voltage 30 Q, if the resistance does not change?

Answer options

  1. 10\,\text{А}
  2. 2{,}5\,\text{А}
  3. 5\,\text{А}
  4. 15\,\text{А}
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Given
U_1
12\,\text{В}
t
10\,\text{с}
q
20\,\text{Кл}
U_2
30\,\text{В}
Find

I_2

  1. I_1=\frac{q}{t}I_1=\frac{20}{10}I_1=2\,\text{А}

    Based on the transferred charge, we determine the initial current.

  2. R=\frac{U_1}{I_1}R=\frac{12}{2}R=6\,\text{Ом}

    We find constant resistance.

  3. I_2=\frac{U_2}{R}I_2=\frac{30}{6}I_2=5\,\text{А}

    We use new voltage.

Answer5\,\text{А}

Position in this set: 17 from 23

Problem #230

Current, resistance and Ohm's law

Under tension 18 V through a resistor for 20 s charge passed 60 Cl. What current will be at voltage 24 Q, if the resistance does not change?

Answer options

  1. 8\,\text{А}
  2. 2\,\text{А}
  3. 12\,\text{А}
  4. 4\,\text{А}
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Given
U_1
18\,\text{В}
t
20\,\text{с}
q
60\,\text{Кл}
U_2
24\,\text{В}
Find

I_2

  1. I_1=\frac{q}{t}I_1=\frac{60}{20}I_1=3\,\text{А}

    Based on the transferred charge, we determine the initial current.

  2. R=\frac{U_1}{I_1}R=\frac{18}{3}R=6\,\text{Ом}

    We find constant resistance.

  3. I_2=\frac{U_2}{R}I_2=\frac{24}{6}I_2=4\,\text{А}

    We use new voltage.

Answer4\,\text{А}

Position in this set: 18 from 23

Problem #231

Electrical work and circuits

Currents flow in two parallel branches 2 and 3 A. Find the current in the unbranched part.

Answer options

  1. 10\,\text{А}
  2. 5\,\text{А}
  3. 2{,}5\,\text{А}
  4. 15\,\text{А}
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Given
I_1
2\,\text{А}
I_2
3\,\text{А}
Find

I

  1. I=I_1+I_2I=2 + 3I=5\,\text{А}

    Charge does not accumulate in the circuit node.

Answer5\,\text{А}

Position in this set: 19 from 23

Problem #232

Electrical work and circuits

Currents flow in two parallel branches 3 and 5 A. Find the current in the unbranched part.

Answer options

  1. 16\,\text{А}
  2. 4\,\text{А}
  3. 8\,\text{А}
  4. 24\,\text{А}
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Given
I_1
3\,\text{А}
I_2
5\,\text{А}
Find

I

  1. I=I_1+I_2I=3 + 5I=8\,\text{А}

    Charge does not accumulate in the circuit node.

Answer8\,\text{А}

Position in this set: 20 from 23

Problem #233

Electrical work and circuits

resistors 6 and 3 Ohms connected in parallel at voltage 12 B. Find the total current.

Answer options

  1. 12\,\text{А}
  2. 3\,\text{А}
  3. 6\,\text{А}
  4. 18\,\text{А}
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Given
R_1
6\,\text{Ом}
R_2
3\,\text{Ом}
U
12\,\text{В}
Find

I

  1. I_1=\frac{U}{R_1}I_1=\frac{12}{6}I_1=2\,\text{А}

    Both branches have the same voltage.

  2. I_2=\frac{U}{R_2}I_2=\frac{12}{3}I_2=4\,\text{А}

    Separately, we find the current of the second branch.

  3. I=I_1+I_2I=2 + 4I=6\,\text{А}

    We add up the branch currents.

Answer6\,\text{А}

Position in this set: 21 from 23

Problem #234

Electrical work and circuits

resistors 8 and 4 Ohms connected in parallel at voltage 24 B. Find the total current.

Answer options

  1. 18\,\text{А}
  2. 4{,}5\,\text{А}
  3. 27\,\text{А}
  4. 9\,\text{А}
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Given
R_1
8\,\text{Ом}
R_2
4\,\text{Ом}
U
24\,\text{В}
Find

I

  1. I_1=\frac{U}{R_1}I_1=\frac{24}{8}I_1=3\,\text{А}

    Both branches have the same voltage.

  2. I_2=\frac{U}{R_2}I_2=\frac{24}{4}I_2=6\,\text{А}

    Separately, we find the current of the second branch.

  3. I=I_1+I_2I=3 + 6I=9\,\text{А}

    We add up the branch currents.

Answer9\,\text{А}

Position in this set: 22 from 23

Problem #235

Electrical work and circuits

resistors 6 and 3 Ohms connected in parallel; there is a resistor in series with this pair 4 Ohm. Whole circuit voltage 18 B. Find the voltage across the parallel pair.

Answer options

  1. 12\,\text{В}
  2. 3\,\text{В}
  3. 18\,\text{В}
  4. 6\,\text{В}
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Given
R_1
6\,\text{Ом}
R_2
3\,\text{Ом}
R_3
4\,\text{Ом}
U
18\,\text{В}
Find

V

  1. R=\frac{R_1R_2}{R_1+R_2}R=\frac{6 \cdot 3}{(6 + 3)}R=2\,\text{Ом}

    We replace the parallel pair with an equivalent resistance.

  2. I=\frac{U}{R+R_3}I=\frac{18}{(2 + 4)}I=3\,\text{А}

    We find the current in the entire circuit.

  3. V=IRV=3 \cdot 2V=6\,\text{В}

    The voltage across the pair is equal to the current of the common circuit multiplied by its resistance.

Answer6\,\text{В}

Position in this set: 23 from 23

Problem #236

Electrical work and circuits

resistors 12 and 6 Ohms connected in parallel; there is a resistor in series with this pair 2 Ohm. Whole circuit voltage 24 B. Find the voltage across the parallel pair.

Answer options

  1. 16\,\text{В}
  2. 32\,\text{В}
  3. 8\,\text{В}
  4. 48\,\text{В}
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Given
R_1
12\,\text{Ом}
R_2
6\,\text{Ом}
R_3
2\,\text{Ом}
U
24\,\text{В}
Find

V

  1. R=\frac{R_1R_2}{R_1+R_2}R=\frac{12 \cdot 6}{(12 + 6)}R=4\,\text{Ом}

    We replace the parallel pair with an equivalent resistance.

  2. I=\frac{U}{R+R_3}I=\frac{24}{(4 + 2)}I=4\,\text{А}

    We find the current in the entire circuit.

  3. V=IRV=4 \cdot 4V=16\,\text{В}

    The voltage across the pair is equal to the current of the common circuit multiplied by its resistance.

Answer16\,\text{В}

The theory behind these problems

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Current, resistance and Ohm's lawCurrent and current density, resistance, emf and Ohm's law.Electrical work and circuitsWork, power, heat, electrolysis and resistor connections.

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Efficiencyη · %PowerP, N · ВтWorkA · ДжElectrical resistanceR · ОмResistivityρ · Ом·мElectromotive forceε · В