152190
A $10 \mathrm{~m}$ long wire of resistance $20 \Omega$ is connected in series with a battery of e.m.f. 3 volt and a resistance of $10 \Omega$. The potential gradient along the wire in volt/meter is
1 0.10
2 0.20
3 0.02
4 1.2
Explanation:
B Using Ohm's law $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {net }}}$ $\mathrm{i}=\frac{3}{10+20}=0.1 \mathrm{~A}$ $\mathrm{V}_{\mathrm{AB}}=0.1 \times 20$ $\mathrm{V}_{\mathrm{AB}}=2 \mathrm{~V}$ Potential gradient, $\lambda=\frac{\mathrm{V}}{\mathrm{l}}$ $\lambda=\frac{2}{10}=0.20 \mathrm{~V} / \mathrm{m}$
MHT-CET 2020
Current Electricity
152192
Using Kirchhoff's law, find the current flowing through the given circuit.
1 $3 \mathrm{~A}$
2 $7.5 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
C Let current in the circuit I Apply KVL in circuit, $\begin{aligned} & -200+38 \mathrm{I}+10=0 \\ & 38 \mathrm{I}=190 \\ & \mathrm{I}=\frac{190}{38}=5 \mathrm{~A} \end{aligned}$
MHT-CET 2020
Current Electricity
152193
In the following network, $I_{1}=-0.4 \mathrm{~A}, I_{4}=1 \mathrm{~A}$ and $I_{5}=0.4$ A. The values of $I_{2}, I_{3}$ and $I_{6}$ respectively are
152190
A $10 \mathrm{~m}$ long wire of resistance $20 \Omega$ is connected in series with a battery of e.m.f. 3 volt and a resistance of $10 \Omega$. The potential gradient along the wire in volt/meter is
1 0.10
2 0.20
3 0.02
4 1.2
Explanation:
B Using Ohm's law $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {net }}}$ $\mathrm{i}=\frac{3}{10+20}=0.1 \mathrm{~A}$ $\mathrm{V}_{\mathrm{AB}}=0.1 \times 20$ $\mathrm{V}_{\mathrm{AB}}=2 \mathrm{~V}$ Potential gradient, $\lambda=\frac{\mathrm{V}}{\mathrm{l}}$ $\lambda=\frac{2}{10}=0.20 \mathrm{~V} / \mathrm{m}$
MHT-CET 2020
Current Electricity
152192
Using Kirchhoff's law, find the current flowing through the given circuit.
1 $3 \mathrm{~A}$
2 $7.5 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
C Let current in the circuit I Apply KVL in circuit, $\begin{aligned} & -200+38 \mathrm{I}+10=0 \\ & 38 \mathrm{I}=190 \\ & \mathrm{I}=\frac{190}{38}=5 \mathrm{~A} \end{aligned}$
MHT-CET 2020
Current Electricity
152193
In the following network, $I_{1}=-0.4 \mathrm{~A}, I_{4}=1 \mathrm{~A}$ and $I_{5}=0.4$ A. The values of $I_{2}, I_{3}$ and $I_{6}$ respectively are
152190
A $10 \mathrm{~m}$ long wire of resistance $20 \Omega$ is connected in series with a battery of e.m.f. 3 volt and a resistance of $10 \Omega$. The potential gradient along the wire in volt/meter is
1 0.10
2 0.20
3 0.02
4 1.2
Explanation:
B Using Ohm's law $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {net }}}$ $\mathrm{i}=\frac{3}{10+20}=0.1 \mathrm{~A}$ $\mathrm{V}_{\mathrm{AB}}=0.1 \times 20$ $\mathrm{V}_{\mathrm{AB}}=2 \mathrm{~V}$ Potential gradient, $\lambda=\frac{\mathrm{V}}{\mathrm{l}}$ $\lambda=\frac{2}{10}=0.20 \mathrm{~V} / \mathrm{m}$
MHT-CET 2020
Current Electricity
152192
Using Kirchhoff's law, find the current flowing through the given circuit.
1 $3 \mathrm{~A}$
2 $7.5 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
C Let current in the circuit I Apply KVL in circuit, $\begin{aligned} & -200+38 \mathrm{I}+10=0 \\ & 38 \mathrm{I}=190 \\ & \mathrm{I}=\frac{190}{38}=5 \mathrm{~A} \end{aligned}$
MHT-CET 2020
Current Electricity
152193
In the following network, $I_{1}=-0.4 \mathrm{~A}, I_{4}=1 \mathrm{~A}$ and $I_{5}=0.4$ A. The values of $I_{2}, I_{3}$ and $I_{6}$ respectively are
152190
A $10 \mathrm{~m}$ long wire of resistance $20 \Omega$ is connected in series with a battery of e.m.f. 3 volt and a resistance of $10 \Omega$. The potential gradient along the wire in volt/meter is
1 0.10
2 0.20
3 0.02
4 1.2
Explanation:
B Using Ohm's law $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\text {net }}}$ $\mathrm{i}=\frac{3}{10+20}=0.1 \mathrm{~A}$ $\mathrm{V}_{\mathrm{AB}}=0.1 \times 20$ $\mathrm{V}_{\mathrm{AB}}=2 \mathrm{~V}$ Potential gradient, $\lambda=\frac{\mathrm{V}}{\mathrm{l}}$ $\lambda=\frac{2}{10}=0.20 \mathrm{~V} / \mathrm{m}$
MHT-CET 2020
Current Electricity
152192
Using Kirchhoff's law, find the current flowing through the given circuit.
1 $3 \mathrm{~A}$
2 $7.5 \mathrm{~A}$
3 $5 \mathrm{~A}$
4 $10 \mathrm{~A}$
Explanation:
C Let current in the circuit I Apply KVL in circuit, $\begin{aligned} & -200+38 \mathrm{I}+10=0 \\ & 38 \mathrm{I}=190 \\ & \mathrm{I}=\frac{190}{38}=5 \mathrm{~A} \end{aligned}$
MHT-CET 2020
Current Electricity
152193
In the following network, $I_{1}=-0.4 \mathrm{~A}, I_{4}=1 \mathrm{~A}$ and $I_{5}=0.4$ A. The values of $I_{2}, I_{3}$ and $I_{6}$ respectively are
