152176
Figure shows a part of an electric circuit. The potentials at points $a, b$ and $c$ are $30 \mathrm{~V}, 12 \mathrm{~V}$ and $2 \mathrm{~V}$ respectively. The current through the $20 \Omega$ resistor will be.
1 $0.4 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $1.0 \mathrm{~A}$
Explanation:
A Given that, $R_{1}=10 \Omega, R_{2}=20 \Omega, R_{3}=30 \Omega$ Apply KCL at point $\mathrm{x}$, $\mathrm{I}_{1}-\mathrm{I}_{2}-\mathrm{I}_{3}=0$ $\frac{30-\mathrm{x}}{10}-\frac{\mathrm{x}-12}{20}-\frac{\mathrm{x}-2}{30}=0$ $\frac{180-6 \mathrm{x}-3 \mathrm{x}+36-2 \mathrm{x}+4}{60}=0$ $220-11 \mathrm{x}=0$ $11 \mathrm{x}=220$ $\mathrm{x}=20 \mathrm{~V}$ Current through the $20 \Omega$ resistor, $=\frac{\mathrm{x}-12}{20}=\left(\frac{20-12}{20}\right)=\frac{8}{20}=\frac{2}{5}=0.4 \text { Amp. }$
JEE Main-06.04.2023
Current Electricity
152177
The equivalent resistance between $A$ and $B$ of the network shown in figure:
152178
As shown in the figure, a network of resistors is connected to battery of $24 \mathrm{~V}$ with an internal resistance of $3 \Omega$. The current through the resistors $R_{4}$ and $R_{5}$ are $I_{4}$ and $I_{5}$ respectively. The values of $I_{4}$ and $I_{5}$ are:
1 $\mathrm{I}_{4}=\frac{8}{5}$ A and $\mathrm{I}_{5}=\frac{2}{5} \mathrm{~A}$
2 $\mathrm{I}_{4}=\frac{6}{5}$ A and $\mathrm{I}_{5}=\frac{24}{5} \mathrm{~A}$
3 $\mathrm{I}_{4}=\frac{2}{5}$ A and $_{5}=\frac{8}{5} \mathrm{~A}$
C Given, $\mathrm{V}=24 \mathrm{~V}$ The equivalent resistance of circuit is- Current through battery is- $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{24}{12}=2 \mathrm{~A}$ Now, $\quad I_{4}=\frac{R_{5}}{R_{4}+R_{5}} \times 2$ $=\frac{5}{20+5} \times 2=\frac{5}{25} \times 2=\frac{2}{5} \mathrm{~A}$ And, $\quad \mathrm{I}_{5}=2-\frac{2}{5}=\frac{10-2}{5}=\frac{8}{5} \mathrm{~A}$ $\therefore \quad \mathrm{I}_{4}=\frac{2}{5} \mathrm{~A}, \mathrm{I}_{5}=\frac{8}{5} \mathrm{~A}$
JEE Main-24.01.2023
Current Electricity
152179
The equivalent resistance between $A$ and B
1 $\frac{2}{3} \Omega$
2 $\frac{3}{2} \Omega$
3 $\frac{1}{3} \Omega$
4 $\frac{1}{2} \Omega$
Explanation:
A The figure for equivalent resistance between $A$ and $B$ can be written as- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{2}+\frac{1}{12}+\frac{1}{4}+\frac{1}{6}+\frac{1}{2}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{6+1+3+2+6}{12}=\frac{18}{12}=\frac{3}{2}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=2 / 3 \Omega$
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Current Electricity
152176
Figure shows a part of an electric circuit. The potentials at points $a, b$ and $c$ are $30 \mathrm{~V}, 12 \mathrm{~V}$ and $2 \mathrm{~V}$ respectively. The current through the $20 \Omega$ resistor will be.
1 $0.4 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $1.0 \mathrm{~A}$
Explanation:
A Given that, $R_{1}=10 \Omega, R_{2}=20 \Omega, R_{3}=30 \Omega$ Apply KCL at point $\mathrm{x}$, $\mathrm{I}_{1}-\mathrm{I}_{2}-\mathrm{I}_{3}=0$ $\frac{30-\mathrm{x}}{10}-\frac{\mathrm{x}-12}{20}-\frac{\mathrm{x}-2}{30}=0$ $\frac{180-6 \mathrm{x}-3 \mathrm{x}+36-2 \mathrm{x}+4}{60}=0$ $220-11 \mathrm{x}=0$ $11 \mathrm{x}=220$ $\mathrm{x}=20 \mathrm{~V}$ Current through the $20 \Omega$ resistor, $=\frac{\mathrm{x}-12}{20}=\left(\frac{20-12}{20}\right)=\frac{8}{20}=\frac{2}{5}=0.4 \text { Amp. }$
JEE Main-06.04.2023
Current Electricity
152177
The equivalent resistance between $A$ and $B$ of the network shown in figure:
152178
As shown in the figure, a network of resistors is connected to battery of $24 \mathrm{~V}$ with an internal resistance of $3 \Omega$. The current through the resistors $R_{4}$ and $R_{5}$ are $I_{4}$ and $I_{5}$ respectively. The values of $I_{4}$ and $I_{5}$ are:
1 $\mathrm{I}_{4}=\frac{8}{5}$ A and $\mathrm{I}_{5}=\frac{2}{5} \mathrm{~A}$
2 $\mathrm{I}_{4}=\frac{6}{5}$ A and $\mathrm{I}_{5}=\frac{24}{5} \mathrm{~A}$
3 $\mathrm{I}_{4}=\frac{2}{5}$ A and $_{5}=\frac{8}{5} \mathrm{~A}$
C Given, $\mathrm{V}=24 \mathrm{~V}$ The equivalent resistance of circuit is- Current through battery is- $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{24}{12}=2 \mathrm{~A}$ Now, $\quad I_{4}=\frac{R_{5}}{R_{4}+R_{5}} \times 2$ $=\frac{5}{20+5} \times 2=\frac{5}{25} \times 2=\frac{2}{5} \mathrm{~A}$ And, $\quad \mathrm{I}_{5}=2-\frac{2}{5}=\frac{10-2}{5}=\frac{8}{5} \mathrm{~A}$ $\therefore \quad \mathrm{I}_{4}=\frac{2}{5} \mathrm{~A}, \mathrm{I}_{5}=\frac{8}{5} \mathrm{~A}$
JEE Main-24.01.2023
Current Electricity
152179
The equivalent resistance between $A$ and B
1 $\frac{2}{3} \Omega$
2 $\frac{3}{2} \Omega$
3 $\frac{1}{3} \Omega$
4 $\frac{1}{2} \Omega$
Explanation:
A The figure for equivalent resistance between $A$ and $B$ can be written as- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{2}+\frac{1}{12}+\frac{1}{4}+\frac{1}{6}+\frac{1}{2}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{6+1+3+2+6}{12}=\frac{18}{12}=\frac{3}{2}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=2 / 3 \Omega$
152176
Figure shows a part of an electric circuit. The potentials at points $a, b$ and $c$ are $30 \mathrm{~V}, 12 \mathrm{~V}$ and $2 \mathrm{~V}$ respectively. The current through the $20 \Omega$ resistor will be.
1 $0.4 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $1.0 \mathrm{~A}$
Explanation:
A Given that, $R_{1}=10 \Omega, R_{2}=20 \Omega, R_{3}=30 \Omega$ Apply KCL at point $\mathrm{x}$, $\mathrm{I}_{1}-\mathrm{I}_{2}-\mathrm{I}_{3}=0$ $\frac{30-\mathrm{x}}{10}-\frac{\mathrm{x}-12}{20}-\frac{\mathrm{x}-2}{30}=0$ $\frac{180-6 \mathrm{x}-3 \mathrm{x}+36-2 \mathrm{x}+4}{60}=0$ $220-11 \mathrm{x}=0$ $11 \mathrm{x}=220$ $\mathrm{x}=20 \mathrm{~V}$ Current through the $20 \Omega$ resistor, $=\frac{\mathrm{x}-12}{20}=\left(\frac{20-12}{20}\right)=\frac{8}{20}=\frac{2}{5}=0.4 \text { Amp. }$
JEE Main-06.04.2023
Current Electricity
152177
The equivalent resistance between $A$ and $B$ of the network shown in figure:
152178
As shown in the figure, a network of resistors is connected to battery of $24 \mathrm{~V}$ with an internal resistance of $3 \Omega$. The current through the resistors $R_{4}$ and $R_{5}$ are $I_{4}$ and $I_{5}$ respectively. The values of $I_{4}$ and $I_{5}$ are:
1 $\mathrm{I}_{4}=\frac{8}{5}$ A and $\mathrm{I}_{5}=\frac{2}{5} \mathrm{~A}$
2 $\mathrm{I}_{4}=\frac{6}{5}$ A and $\mathrm{I}_{5}=\frac{24}{5} \mathrm{~A}$
3 $\mathrm{I}_{4}=\frac{2}{5}$ A and $_{5}=\frac{8}{5} \mathrm{~A}$
C Given, $\mathrm{V}=24 \mathrm{~V}$ The equivalent resistance of circuit is- Current through battery is- $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{24}{12}=2 \mathrm{~A}$ Now, $\quad I_{4}=\frac{R_{5}}{R_{4}+R_{5}} \times 2$ $=\frac{5}{20+5} \times 2=\frac{5}{25} \times 2=\frac{2}{5} \mathrm{~A}$ And, $\quad \mathrm{I}_{5}=2-\frac{2}{5}=\frac{10-2}{5}=\frac{8}{5} \mathrm{~A}$ $\therefore \quad \mathrm{I}_{4}=\frac{2}{5} \mathrm{~A}, \mathrm{I}_{5}=\frac{8}{5} \mathrm{~A}$
JEE Main-24.01.2023
Current Electricity
152179
The equivalent resistance between $A$ and B
1 $\frac{2}{3} \Omega$
2 $\frac{3}{2} \Omega$
3 $\frac{1}{3} \Omega$
4 $\frac{1}{2} \Omega$
Explanation:
A The figure for equivalent resistance between $A$ and $B$ can be written as- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{2}+\frac{1}{12}+\frac{1}{4}+\frac{1}{6}+\frac{1}{2}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{6+1+3+2+6}{12}=\frac{18}{12}=\frac{3}{2}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=2 / 3 \Omega$
152176
Figure shows a part of an electric circuit. The potentials at points $a, b$ and $c$ are $30 \mathrm{~V}, 12 \mathrm{~V}$ and $2 \mathrm{~V}$ respectively. The current through the $20 \Omega$ resistor will be.
1 $0.4 \mathrm{~A}$
2 $0.2 \mathrm{~A}$
3 $0.6 \mathrm{~A}$
4 $1.0 \mathrm{~A}$
Explanation:
A Given that, $R_{1}=10 \Omega, R_{2}=20 \Omega, R_{3}=30 \Omega$ Apply KCL at point $\mathrm{x}$, $\mathrm{I}_{1}-\mathrm{I}_{2}-\mathrm{I}_{3}=0$ $\frac{30-\mathrm{x}}{10}-\frac{\mathrm{x}-12}{20}-\frac{\mathrm{x}-2}{30}=0$ $\frac{180-6 \mathrm{x}-3 \mathrm{x}+36-2 \mathrm{x}+4}{60}=0$ $220-11 \mathrm{x}=0$ $11 \mathrm{x}=220$ $\mathrm{x}=20 \mathrm{~V}$ Current through the $20 \Omega$ resistor, $=\frac{\mathrm{x}-12}{20}=\left(\frac{20-12}{20}\right)=\frac{8}{20}=\frac{2}{5}=0.4 \text { Amp. }$
JEE Main-06.04.2023
Current Electricity
152177
The equivalent resistance between $A$ and $B$ of the network shown in figure:
152178
As shown in the figure, a network of resistors is connected to battery of $24 \mathrm{~V}$ with an internal resistance of $3 \Omega$. The current through the resistors $R_{4}$ and $R_{5}$ are $I_{4}$ and $I_{5}$ respectively. The values of $I_{4}$ and $I_{5}$ are:
1 $\mathrm{I}_{4}=\frac{8}{5}$ A and $\mathrm{I}_{5}=\frac{2}{5} \mathrm{~A}$
2 $\mathrm{I}_{4}=\frac{6}{5}$ A and $\mathrm{I}_{5}=\frac{24}{5} \mathrm{~A}$
3 $\mathrm{I}_{4}=\frac{2}{5}$ A and $_{5}=\frac{8}{5} \mathrm{~A}$
C Given, $\mathrm{V}=24 \mathrm{~V}$ The equivalent resistance of circuit is- Current through battery is- $\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{24}{12}=2 \mathrm{~A}$ Now, $\quad I_{4}=\frac{R_{5}}{R_{4}+R_{5}} \times 2$ $=\frac{5}{20+5} \times 2=\frac{5}{25} \times 2=\frac{2}{5} \mathrm{~A}$ And, $\quad \mathrm{I}_{5}=2-\frac{2}{5}=\frac{10-2}{5}=\frac{8}{5} \mathrm{~A}$ $\therefore \quad \mathrm{I}_{4}=\frac{2}{5} \mathrm{~A}, \mathrm{I}_{5}=\frac{8}{5} \mathrm{~A}$
JEE Main-24.01.2023
Current Electricity
152179
The equivalent resistance between $A$ and B
1 $\frac{2}{3} \Omega$
2 $\frac{3}{2} \Omega$
3 $\frac{1}{3} \Omega$
4 $\frac{1}{2} \Omega$
Explanation:
A The figure for equivalent resistance between $A$ and $B$ can be written as- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{2}+\frac{1}{12}+\frac{1}{4}+\frac{1}{6}+\frac{1}{2}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{6+1+3+2+6}{12}=\frac{18}{12}=\frac{3}{2}$ $\therefore \quad \mathrm{R}_{\mathrm{eq}}=2 / 3 \Omega$
