GRADE 8 PRACTICE
Physics Problems: Resistance and Ohm's Law, Grade 8
Conductor resistance, Ohm's law and rheostat.
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
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Problem #43
Resistance and Ohm's LawWhat concept does the description correspond to: “One ohm is the resistance of the section in which a voltage of one volt creates a current of one ampere.»?
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- \text{Description}
- \text{One ohm is the resistance of the section in which a voltage of one volt creates a current of one ampere.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
The unit relates the readings of the ammeter and voltmeter through Ohm's law.
- \text{Let's formulate conclusion}\text{Concept} = \text{Ом}
Answer\text{Concept} = \text{Ом}
Problem #44
Resistance and Ohm's LawWhat concept does the description correspond to: “For an ohmic section under constant physical conditions, current is proportional to voltage: I = U/R, where R is constant over the range considered.»?
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- \text{Description}
- \text{For an ohmic section under constant physical conditions, current is proportional to voltage: I = U/R, where R is constant over the range considered.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
The graph of I(U) of such a section is a straight line through the origin. Heating can change R. However, constant temperature alone is not enough: the diode and some other elements remain non-ohmic, so the applicability of the dependence with constant R is tested by experiment.
- \text{Let's formulate conclusion}\text{Concept} = \text{Ohm's law for a circuit section}
Answer\text{Concept} = \text{Ohm's law for a circuit section}
Problem #45
Ohm's lawA 6 ohm resistor maintains a voltage of 12 V. Find the current.
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- R
- 6\,\text{Ом}
- U
- 12\,\text{В}
I - ?
- I=\frac{U}{R}I=\frac{12}{6}\,\text{А}I = 2\,\text{А}
Answer2\,\text{А}
Problem #46
Conductor resistance and rheostatWhat concept does the description correspond to: “The resistance of a homogeneous conductor increases with length, decreases with cross-sectional area and depends on the material.»?
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- \text{Description}
- \text{The resistance of a homogeneous conductor increases with length, decreases with cross-sectional area and depends on the material.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
A long, thin wire has more resistance than a short, thick wire of the same material.
- \text{Let's formulate conclusion}\text{Concept} = \text{Conductor resistance}
Answer\text{Concept} = \text{Conductor resistance}
Problem #47
Conductor resistance and rheostatWhat concept does the description correspond to: “A characteristic of a material that relates the resistance of a conductor to its length and cross-sectional area.»?
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- \text{Description}
- \text{A characteristic of a material that relates the resistance of a conductor to its length and cross-sectional area.}
\text{name concepts} - ?
- \text{Let's compare approval with definition}
Copper's low resistivity makes it useful for wiring; alloys with greater resistivity are useful for heaters.
- \text{Let's formulate conclusion}\text{Concept} = \text{Resistivity}
Answer\text{Concept} = \text{Resistivity}
Problem #48
Conductor resistanceA wire 10 m long and with a cross-sectional area of 2 mm² has a resistivity of 0.5 Ohm mm²/m. Find resistance.
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- \rho
- 0{,}5\,\text{Ом}\cdot\text{мм}^{2}/\text{м}
- l
- 10\,\text{м}
- S
- 2\,\text{мм}^{2}
R - ?
- R=\rho\frac{l}{S}R=0{,}5\cdot\frac{10}{2}\,\text{Ом}R = 2{,}5\,\text{Ом}
Answer2{,}5\,\text{Ом}
Problem #157
Resistance and Ohm's LawResistor voltage 12 V, resistance 4 Ohm. Find the current.
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- U
- 12\,\text{В}
- R
- 4\,\text{Ом}
I - ?
- 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}
Problem #158
Resistance and Ohm's LawResistor voltage 24 V, resistance 6 Ohm. Find the current.
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- U
- 24\,\text{В}
- R
- 6\,\text{Ом}
I - ?
- 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}
Problem #159
Resistance and Ohm's LawUnder tension 6 The ammeter shows 200 mA. Find the resistance of the circuit section.
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- U
- 6\,\text{В}
- i
- 200\,\text{мА}
R - ?
- I=\frac{i}{1000}I=\frac{200}{1000}I=0{,}2\,\text{А}
Before calculating, we convert milliamps to amperes.
- R=\frac{U}{I}R=\frac{6}{0{,}2}R=30\,\text{Ом}
We express resistance from Ohm's law.
Answer30\,\text{Ом}
Problem #160
Resistance and Ohm's LawUnder tension 12 The ammeter shows 300 mA. Find the resistance of the circuit section.
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- U
- 12\,\text{В}
- i
- 300\,\text{мА}
R - ?
- I=\frac{i}{1000}I=\frac{300}{1000}I=0{,}3\,\text{А}
Before calculating, we convert milliamps to amperes.
- R=\frac{U}{I}R=\frac{12}{0{,}3}R=40\,\text{Ом}
We express resistance from Ohm's law.
Answer40\,\text{Ом}
Problem #161
Resistance and Ohm's LawUnder 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?
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- U_1
- 12\,\text{В}
- t
- 10\,\text{с}
- q
- 20\,\text{Кл}
- U_2
- 30\,\text{В}
I_2 - ?
- 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.
- R=\frac{U_1}{I_1}R=\frac{12}{2}R=6\,\text{Ом}
We find constant resistance.
- I_2=\frac{U_2}{R}I_2=\frac{30}{6}I_2=5\,\text{А}
We use new voltage.
Answer5\,\text{А}
Problem #162
Resistance and Ohm's LawUnder 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?
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- U_1
- 18\,\text{В}
- t
- 20\,\text{с}
- q
- 60\,\text{Кл}
- U_2
- 24\,\text{В}
I_2 - ?
- 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.
- R=\frac{U_1}{I_1}R=\frac{18}{3}R=6\,\text{Ом}
We find constant resistance.
- I_2=\frac{U_2}{R}I_2=\frac{24}{6}I_2=4\,\text{А}
We use new voltage.
Answer4\,\text{А}
Problem #163
Conductor resistance and rheostatWire length 10 m andmherт section 0{,}5 mm². Resistivity 0{,}1 Ohm mm²/m. Find resistance.
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- l
- 10\,\text{м}
- S
- 0{,}5\,\text{мм}^{2}
- r
- 0{,}1\,\text{Ом·мм}^{2}\text{/м}
R - ?
- R=\rho\frac{l}{S}R=\frac{0{,}1 \cdot 10}{0{,}5}R=2\,\text{Ом}
Resistivity units are consistent with length and cross-section.
Answer2\,\text{Ом}
Problem #164
Conductor resistance and rheostatWire length 20 m andmherт section 0{,}4 mm². Resistivity 0{,}02 Ohm mm²/m. Find resistance.
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- l
- 20\,\text{м}
- S
- 0{,}4\,\text{мм}^{2}
- r
- 0{,}02\,\text{Ом·мм}^{2}\text{/м}
R - ?
- R=\rho\frac{l}{S}R=\frac{0{,}02 \cdot 20}{0{,}4}R=1\,\text{Ом}
Resistivity units are consistent with length and cross-section.
Answer1\,\text{Ом}
Problem #165
Conductor resistance and rheostatFull wire rheostat length 10 m andmherт resistance 20 Ohm. slider includes B chain 3 m wire. Find force current Under tension 12 B on this parts.
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- l
- 10\,\text{м}
- R
- 20\,\text{Ом}
- a
- 3\,\text{м}
- U
- 12\,\text{В}
I - ?
- r=R\frac{a}{l}r=\frac{20 \cdot 3}{10}r=6\,\text{Ом}
The resistance of a homogeneous wire is proportional to its length.
- I=\frac{U}{r}I=\frac{12}{6}I=2\,\text{А}
Ohm's law applies only to the included part.
Answer2\,\text{А}
Problem #166
Conductor resistance and rheostatFull wire rheostat length 12 m andmherт resistance 24 Ohm. slider includes B chain 4 m wire. Find force current Under tension 24 B on this parts.
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- l
- 12\,\text{м}
- R
- 24\,\text{Ом}
- a
- 4\,\text{м}
- U
- 24\,\text{В}
I - ?
- r=R\frac{a}{l}r=\frac{24 \cdot 4}{12}r=8\,\text{Ом}
The resistance of a homogeneous wire is proportional to its length.
- I=\frac{U}{r}I=\frac{24}{8}I=3\,\text{А}
Ohm's law applies only to the included part.
Answer3\,\text{А}
Problem #167
Conductor resistance and rheostatWire with specific resistance 0{,}5 Ohm mm²/m and cross section 0{,}2 mm² should skip current 2 A Under tension 10 B. Find mass this wire, if her density 8 g/cm³. Please note: 1 m·mm² = 1 cm³.
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- r
- 0{,}5\,\text{Ом·мм}^{2}\text{/м}
- S
- 0{,}2\,\text{мм}^{2}
- I
- 2\,\text{А}
- U
- 10\,\text{В}
- d
- 8\,\text{г/см}^{3}
m - ?
- R=\frac{U}{I}R=\frac{10}{2}R=5\,\text{Ом}
First we determine the required resistance.
- l=\frac{RS}{\rho}l=\frac{5 \cdot 0{,}2}{0{,}5}l=2\,\text{м}
Express the length of the wire.
- V=lSV=2 \cdot 0{,}2V=0{,}4\,\text{см}^{3}
We use the correspondence of units specified in the condition.
- m=dVm=8 \cdot 0{,}4m=3{,}2\,\text{г}
We find mass by density and volume.
Answer3{,}2\,\text{г}
Problem #168
Conductor resistance and rheostatWire with specific resistance 0{,}4 Ohm mm²/m and cross section 0{,}3 mm² should skip current 3 A Under tension 12 B. Find mass this wire, if her density 8 g/cm³. Please note: 1 m·mm² = 1 cm³.
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- r
- 0{,}4\,\text{Ом·мм}^{2}\text{/м}
- S
- 0{,}3\,\text{мм}^{2}
- I
- 3\,\text{А}
- U
- 12\,\text{В}
- d
- 8\,\text{г/см}^{3}
m - ?
- R=\frac{U}{I}R=\frac{12}{3}R=4\,\text{Ом}
First we determine the required resistance.
- l=\frac{RS}{\rho}l=\frac{4 \cdot 0{,}3}{0{,}4}l=3\,\text{м}
Express the length of the wire.
- V=lSV=3 \cdot 0{,}3V=0{,}9\,\text{см}^{3}
We use the correspondence of units specified in the condition.
- m=dVm=8 \cdot 0{,}9m=7{,}2\,\text{г}
We find mass by density and volume.
Answer7{,}2\,\text{г}
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