152171
The equivalent resistance between $A$ and $B$ as shown in figure is :
1 $5 \mathrm{k} \Omega$
2 $30 \mathrm{k} \Omega$
3 $10 \mathrm{k} \Omega$
4 $20 \mathrm{k} \Omega$
Explanation:
A Given circuit, $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}=0$ $\mathrm{~V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}}$ All resistors are in parallel. So, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}$ $\mathrm{R}_{\mathrm{eq}}=5 \mathrm{k} \Omega$
JEE Main-08.04.2023
Current Electricity
152172
The equivalent resistance of the circuit shown below between points a and $b$ is:
1 $24 \Omega$
2 $3.2 \Omega$
3 $20 \Omega$
4 $16 \Omega$
Explanation:
B The circuit can be reduced to $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{2+1+2}{16}=\frac{5}{16}$ $\mathrm{R}_{\mathrm{eq}}=\frac{16}{5}=3.2 \Omega$
JEE Main-10.04.2023
Current Electricity
152173
In the given circuit, the current (I) through the battery will be:
1 $1.5 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2.5 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A In the circuit $D_{1}$ and $D_{3}$ are forward biased and $\mathrm{D}_{2}$ is reverse biased. We know that, Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}$ $\therefore \quad \mathrm{I}=\frac{10}{20 / 3}=\frac{3}{2} \mathrm{~A}=1.5 \mathrm{~A}$
JEE Main-15.04.2023
Current Electricity
152174
In this figure the resistance of the coil of galvanometer $G$ is $2 \Omega$. The emf of the cell is 4 $V$. The ratio of potential difference across $C_{1}$ and $\mathrm{C}_{2}$ is :
1 1
2 $\frac{4}{5}$
3 $\frac{3}{4}$
4 $\frac{5}{4}$
Explanation:
B At steady condition, Current through capacitor $=0$ $\mathrm{R}_{\mathrm{eq}}=6 \Omega+2 \Omega+8 \Omega=16 \Omega$ We know that, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{4 \mathrm{~V}}{16 \Omega}=\frac{1}{4} \mathrm{~A}$ Now, Voltage across $\mathrm{C}_{1}$, $\mathrm{V}_{\mathrm{AC}}=\mathrm{I}(6 \Omega+2 \Omega)$ $\mathrm{V}_{\mathrm{AC}}=\frac{1}{4} \times 8=2 \mathrm{~V}$ And Voltage across $\mathrm{C}_{2}$, $\mathrm{V}_{\mathrm{BD}}=\mathrm{I}(2 \Omega+8 \Omega)$ $\mathrm{V}_{\mathrm{BD}}=\frac{1}{4} \times 10=\frac{5}{2} \mathrm{~V}$ On dividing equation (i) and (ii), we get- $\frac{\mathrm{V}_{\mathrm{AC}}}{\mathrm{V}_{\mathrm{BD}}}=\frac{2}{\frac{5}{2}}=\frac{4}{5}$ Hence, the ratio of potential difference across $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is $4: 5$.
JEE Main-08.04.2023
Current Electricity
152175
Different combination of 3 resistors of equal resistance $R$ are shown in the figures. The increasing order for power dissipation is : $\mathrm{R}$
D We know that, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ Since, $\mathrm{I}$ is same. So, $\mathrm{P} \propto \mathrm{R}$. For (A): $\therefore$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{2}+\mathrm{R}=\frac{3}{2} \mathrm{R}$ $\therefore$ For (B): $\mathrm{P}_{\mathrm{A}}=\mathrm{I}^{2}\left(\frac{3 \mathrm{R}}{2}\right)$ $\mathrm{P}_{\mathrm{B}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{2 \mathrm{R}}+\frac{1}{\mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ $\therefore \quad \mathrm{P}_{\mathrm{B}}=\mathrm{I}^{2}\left(\frac{2 \mathrm{R}}{3}\right)$ For $(C): \mathrm{P}_{\mathrm{C}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{3} \quad \mathrm{P}_{\mathrm{C}}=\mathrm{I}^{2}\left(\frac{\mathrm{R}}{3}\right)$ For (D): $\text { : } P_{P^{2}=I^{2} R_{e q}}^{R}$ Now, increasing order of power dissipation is, $\mathrm{P}_{\mathrm{C}}\lt\mathrm{P}_{\mathrm{B}}\lt\mathrm{P}_{\mathrm{A}}\lt\mathrm{P}_{\mathrm{D}}$
152171
The equivalent resistance between $A$ and $B$ as shown in figure is :
1 $5 \mathrm{k} \Omega$
2 $30 \mathrm{k} \Omega$
3 $10 \mathrm{k} \Omega$
4 $20 \mathrm{k} \Omega$
Explanation:
A Given circuit, $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}=0$ $\mathrm{~V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}}$ All resistors are in parallel. So, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}$ $\mathrm{R}_{\mathrm{eq}}=5 \mathrm{k} \Omega$
JEE Main-08.04.2023
Current Electricity
152172
The equivalent resistance of the circuit shown below between points a and $b$ is:
1 $24 \Omega$
2 $3.2 \Omega$
3 $20 \Omega$
4 $16 \Omega$
Explanation:
B The circuit can be reduced to $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{2+1+2}{16}=\frac{5}{16}$ $\mathrm{R}_{\mathrm{eq}}=\frac{16}{5}=3.2 \Omega$
JEE Main-10.04.2023
Current Electricity
152173
In the given circuit, the current (I) through the battery will be:
1 $1.5 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2.5 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A In the circuit $D_{1}$ and $D_{3}$ are forward biased and $\mathrm{D}_{2}$ is reverse biased. We know that, Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}$ $\therefore \quad \mathrm{I}=\frac{10}{20 / 3}=\frac{3}{2} \mathrm{~A}=1.5 \mathrm{~A}$
JEE Main-15.04.2023
Current Electricity
152174
In this figure the resistance of the coil of galvanometer $G$ is $2 \Omega$. The emf of the cell is 4 $V$. The ratio of potential difference across $C_{1}$ and $\mathrm{C}_{2}$ is :
1 1
2 $\frac{4}{5}$
3 $\frac{3}{4}$
4 $\frac{5}{4}$
Explanation:
B At steady condition, Current through capacitor $=0$ $\mathrm{R}_{\mathrm{eq}}=6 \Omega+2 \Omega+8 \Omega=16 \Omega$ We know that, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{4 \mathrm{~V}}{16 \Omega}=\frac{1}{4} \mathrm{~A}$ Now, Voltage across $\mathrm{C}_{1}$, $\mathrm{V}_{\mathrm{AC}}=\mathrm{I}(6 \Omega+2 \Omega)$ $\mathrm{V}_{\mathrm{AC}}=\frac{1}{4} \times 8=2 \mathrm{~V}$ And Voltage across $\mathrm{C}_{2}$, $\mathrm{V}_{\mathrm{BD}}=\mathrm{I}(2 \Omega+8 \Omega)$ $\mathrm{V}_{\mathrm{BD}}=\frac{1}{4} \times 10=\frac{5}{2} \mathrm{~V}$ On dividing equation (i) and (ii), we get- $\frac{\mathrm{V}_{\mathrm{AC}}}{\mathrm{V}_{\mathrm{BD}}}=\frac{2}{\frac{5}{2}}=\frac{4}{5}$ Hence, the ratio of potential difference across $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is $4: 5$.
JEE Main-08.04.2023
Current Electricity
152175
Different combination of 3 resistors of equal resistance $R$ are shown in the figures. The increasing order for power dissipation is : $\mathrm{R}$
D We know that, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ Since, $\mathrm{I}$ is same. So, $\mathrm{P} \propto \mathrm{R}$. For (A): $\therefore$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{2}+\mathrm{R}=\frac{3}{2} \mathrm{R}$ $\therefore$ For (B): $\mathrm{P}_{\mathrm{A}}=\mathrm{I}^{2}\left(\frac{3 \mathrm{R}}{2}\right)$ $\mathrm{P}_{\mathrm{B}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{2 \mathrm{R}}+\frac{1}{\mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ $\therefore \quad \mathrm{P}_{\mathrm{B}}=\mathrm{I}^{2}\left(\frac{2 \mathrm{R}}{3}\right)$ For $(C): \mathrm{P}_{\mathrm{C}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{3} \quad \mathrm{P}_{\mathrm{C}}=\mathrm{I}^{2}\left(\frac{\mathrm{R}}{3}\right)$ For (D): $\text { : } P_{P^{2}=I^{2} R_{e q}}^{R}$ Now, increasing order of power dissipation is, $\mathrm{P}_{\mathrm{C}}\lt\mathrm{P}_{\mathrm{B}}\lt\mathrm{P}_{\mathrm{A}}\lt\mathrm{P}_{\mathrm{D}}$
152171
The equivalent resistance between $A$ and $B$ as shown in figure is :
1 $5 \mathrm{k} \Omega$
2 $30 \mathrm{k} \Omega$
3 $10 \mathrm{k} \Omega$
4 $20 \mathrm{k} \Omega$
Explanation:
A Given circuit, $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}=0$ $\mathrm{~V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}}$ All resistors are in parallel. So, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}$ $\mathrm{R}_{\mathrm{eq}}=5 \mathrm{k} \Omega$
JEE Main-08.04.2023
Current Electricity
152172
The equivalent resistance of the circuit shown below between points a and $b$ is:
1 $24 \Omega$
2 $3.2 \Omega$
3 $20 \Omega$
4 $16 \Omega$
Explanation:
B The circuit can be reduced to $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{2+1+2}{16}=\frac{5}{16}$ $\mathrm{R}_{\mathrm{eq}}=\frac{16}{5}=3.2 \Omega$
JEE Main-10.04.2023
Current Electricity
152173
In the given circuit, the current (I) through the battery will be:
1 $1.5 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2.5 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A In the circuit $D_{1}$ and $D_{3}$ are forward biased and $\mathrm{D}_{2}$ is reverse biased. We know that, Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}$ $\therefore \quad \mathrm{I}=\frac{10}{20 / 3}=\frac{3}{2} \mathrm{~A}=1.5 \mathrm{~A}$
JEE Main-15.04.2023
Current Electricity
152174
In this figure the resistance of the coil of galvanometer $G$ is $2 \Omega$. The emf of the cell is 4 $V$. The ratio of potential difference across $C_{1}$ and $\mathrm{C}_{2}$ is :
1 1
2 $\frac{4}{5}$
3 $\frac{3}{4}$
4 $\frac{5}{4}$
Explanation:
B At steady condition, Current through capacitor $=0$ $\mathrm{R}_{\mathrm{eq}}=6 \Omega+2 \Omega+8 \Omega=16 \Omega$ We know that, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{4 \mathrm{~V}}{16 \Omega}=\frac{1}{4} \mathrm{~A}$ Now, Voltage across $\mathrm{C}_{1}$, $\mathrm{V}_{\mathrm{AC}}=\mathrm{I}(6 \Omega+2 \Omega)$ $\mathrm{V}_{\mathrm{AC}}=\frac{1}{4} \times 8=2 \mathrm{~V}$ And Voltage across $\mathrm{C}_{2}$, $\mathrm{V}_{\mathrm{BD}}=\mathrm{I}(2 \Omega+8 \Omega)$ $\mathrm{V}_{\mathrm{BD}}=\frac{1}{4} \times 10=\frac{5}{2} \mathrm{~V}$ On dividing equation (i) and (ii), we get- $\frac{\mathrm{V}_{\mathrm{AC}}}{\mathrm{V}_{\mathrm{BD}}}=\frac{2}{\frac{5}{2}}=\frac{4}{5}$ Hence, the ratio of potential difference across $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is $4: 5$.
JEE Main-08.04.2023
Current Electricity
152175
Different combination of 3 resistors of equal resistance $R$ are shown in the figures. The increasing order for power dissipation is : $\mathrm{R}$
D We know that, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ Since, $\mathrm{I}$ is same. So, $\mathrm{P} \propto \mathrm{R}$. For (A): $\therefore$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{2}+\mathrm{R}=\frac{3}{2} \mathrm{R}$ $\therefore$ For (B): $\mathrm{P}_{\mathrm{A}}=\mathrm{I}^{2}\left(\frac{3 \mathrm{R}}{2}\right)$ $\mathrm{P}_{\mathrm{B}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{2 \mathrm{R}}+\frac{1}{\mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ $\therefore \quad \mathrm{P}_{\mathrm{B}}=\mathrm{I}^{2}\left(\frac{2 \mathrm{R}}{3}\right)$ For $(C): \mathrm{P}_{\mathrm{C}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{3} \quad \mathrm{P}_{\mathrm{C}}=\mathrm{I}^{2}\left(\frac{\mathrm{R}}{3}\right)$ For (D): $\text { : } P_{P^{2}=I^{2} R_{e q}}^{R}$ Now, increasing order of power dissipation is, $\mathrm{P}_{\mathrm{C}}\lt\mathrm{P}_{\mathrm{B}}\lt\mathrm{P}_{\mathrm{A}}\lt\mathrm{P}_{\mathrm{D}}$
152171
The equivalent resistance between $A$ and $B$ as shown in figure is :
1 $5 \mathrm{k} \Omega$
2 $30 \mathrm{k} \Omega$
3 $10 \mathrm{k} \Omega$
4 $20 \mathrm{k} \Omega$
Explanation:
A Given circuit, $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}=0$ $\mathrm{~V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}}$ All resistors are in parallel. So, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}$ $\mathrm{R}_{\mathrm{eq}}=5 \mathrm{k} \Omega$
JEE Main-08.04.2023
Current Electricity
152172
The equivalent resistance of the circuit shown below between points a and $b$ is:
1 $24 \Omega$
2 $3.2 \Omega$
3 $20 \Omega$
4 $16 \Omega$
Explanation:
B The circuit can be reduced to $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{2+1+2}{16}=\frac{5}{16}$ $\mathrm{R}_{\mathrm{eq}}=\frac{16}{5}=3.2 \Omega$
JEE Main-10.04.2023
Current Electricity
152173
In the given circuit, the current (I) through the battery will be:
1 $1.5 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2.5 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A In the circuit $D_{1}$ and $D_{3}$ are forward biased and $\mathrm{D}_{2}$ is reverse biased. We know that, Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}$ $\therefore \quad \mathrm{I}=\frac{10}{20 / 3}=\frac{3}{2} \mathrm{~A}=1.5 \mathrm{~A}$
JEE Main-15.04.2023
Current Electricity
152174
In this figure the resistance of the coil of galvanometer $G$ is $2 \Omega$. The emf of the cell is 4 $V$. The ratio of potential difference across $C_{1}$ and $\mathrm{C}_{2}$ is :
1 1
2 $\frac{4}{5}$
3 $\frac{3}{4}$
4 $\frac{5}{4}$
Explanation:
B At steady condition, Current through capacitor $=0$ $\mathrm{R}_{\mathrm{eq}}=6 \Omega+2 \Omega+8 \Omega=16 \Omega$ We know that, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{4 \mathrm{~V}}{16 \Omega}=\frac{1}{4} \mathrm{~A}$ Now, Voltage across $\mathrm{C}_{1}$, $\mathrm{V}_{\mathrm{AC}}=\mathrm{I}(6 \Omega+2 \Omega)$ $\mathrm{V}_{\mathrm{AC}}=\frac{1}{4} \times 8=2 \mathrm{~V}$ And Voltage across $\mathrm{C}_{2}$, $\mathrm{V}_{\mathrm{BD}}=\mathrm{I}(2 \Omega+8 \Omega)$ $\mathrm{V}_{\mathrm{BD}}=\frac{1}{4} \times 10=\frac{5}{2} \mathrm{~V}$ On dividing equation (i) and (ii), we get- $\frac{\mathrm{V}_{\mathrm{AC}}}{\mathrm{V}_{\mathrm{BD}}}=\frac{2}{\frac{5}{2}}=\frac{4}{5}$ Hence, the ratio of potential difference across $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is $4: 5$.
JEE Main-08.04.2023
Current Electricity
152175
Different combination of 3 resistors of equal resistance $R$ are shown in the figures. The increasing order for power dissipation is : $\mathrm{R}$
D We know that, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ Since, $\mathrm{I}$ is same. So, $\mathrm{P} \propto \mathrm{R}$. For (A): $\therefore$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{2}+\mathrm{R}=\frac{3}{2} \mathrm{R}$ $\therefore$ For (B): $\mathrm{P}_{\mathrm{A}}=\mathrm{I}^{2}\left(\frac{3 \mathrm{R}}{2}\right)$ $\mathrm{P}_{\mathrm{B}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{2 \mathrm{R}}+\frac{1}{\mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ $\therefore \quad \mathrm{P}_{\mathrm{B}}=\mathrm{I}^{2}\left(\frac{2 \mathrm{R}}{3}\right)$ For $(C): \mathrm{P}_{\mathrm{C}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{3} \quad \mathrm{P}_{\mathrm{C}}=\mathrm{I}^{2}\left(\frac{\mathrm{R}}{3}\right)$ For (D): $\text { : } P_{P^{2}=I^{2} R_{e q}}^{R}$ Now, increasing order of power dissipation is, $\mathrm{P}_{\mathrm{C}}\lt\mathrm{P}_{\mathrm{B}}\lt\mathrm{P}_{\mathrm{A}}\lt\mathrm{P}_{\mathrm{D}}$
152171
The equivalent resistance between $A$ and $B$ as shown in figure is :
1 $5 \mathrm{k} \Omega$
2 $30 \mathrm{k} \Omega$
3 $10 \mathrm{k} \Omega$
4 $20 \mathrm{k} \Omega$
Explanation:
A Given circuit, $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}=0$ $\mathrm{~V}_{\mathrm{C}}=\mathrm{V}_{\mathrm{B}}$ All resistors are in parallel. So, $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{20}$ $\mathrm{R}_{\mathrm{eq}}=5 \mathrm{k} \Omega$
JEE Main-08.04.2023
Current Electricity
152172
The equivalent resistance of the circuit shown below between points a and $b$ is:
1 $24 \Omega$
2 $3.2 \Omega$
3 $20 \Omega$
4 $16 \Omega$
Explanation:
B The circuit can be reduced to $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{2+1+2}{16}=\frac{5}{16}$ $\mathrm{R}_{\mathrm{eq}}=\frac{16}{5}=3.2 \Omega$
JEE Main-10.04.2023
Current Electricity
152173
In the given circuit, the current (I) through the battery will be:
1 $1.5 \mathrm{~A}$
2 $1 \mathrm{~A}$
3 $2.5 \mathrm{~A}$
4 $2 \mathrm{~A}$
Explanation:
A In the circuit $D_{1}$ and $D_{3}$ are forward biased and $\mathrm{D}_{2}$ is reverse biased. We know that, Current $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}$ $\therefore \quad \mathrm{I}=\frac{10}{20 / 3}=\frac{3}{2} \mathrm{~A}=1.5 \mathrm{~A}$
JEE Main-15.04.2023
Current Electricity
152174
In this figure the resistance of the coil of galvanometer $G$ is $2 \Omega$. The emf of the cell is 4 $V$. The ratio of potential difference across $C_{1}$ and $\mathrm{C}_{2}$ is :
1 1
2 $\frac{4}{5}$
3 $\frac{3}{4}$
4 $\frac{5}{4}$
Explanation:
B At steady condition, Current through capacitor $=0$ $\mathrm{R}_{\mathrm{eq}}=6 \Omega+2 \Omega+8 \Omega=16 \Omega$ We know that, $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{4 \mathrm{~V}}{16 \Omega}=\frac{1}{4} \mathrm{~A}$ Now, Voltage across $\mathrm{C}_{1}$, $\mathrm{V}_{\mathrm{AC}}=\mathrm{I}(6 \Omega+2 \Omega)$ $\mathrm{V}_{\mathrm{AC}}=\frac{1}{4} \times 8=2 \mathrm{~V}$ And Voltage across $\mathrm{C}_{2}$, $\mathrm{V}_{\mathrm{BD}}=\mathrm{I}(2 \Omega+8 \Omega)$ $\mathrm{V}_{\mathrm{BD}}=\frac{1}{4} \times 10=\frac{5}{2} \mathrm{~V}$ On dividing equation (i) and (ii), we get- $\frac{\mathrm{V}_{\mathrm{AC}}}{\mathrm{V}_{\mathrm{BD}}}=\frac{2}{\frac{5}{2}}=\frac{4}{5}$ Hence, the ratio of potential difference across $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ is $4: 5$.
JEE Main-08.04.2023
Current Electricity
152175
Different combination of 3 resistors of equal resistance $R$ are shown in the figures. The increasing order for power dissipation is : $\mathrm{R}$
D We know that, $\mathrm{P}=\mathrm{I}^{2} \mathrm{R}$ Since, $\mathrm{I}$ is same. So, $\mathrm{P} \propto \mathrm{R}$. For (A): $\therefore$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{2}+\mathrm{R}=\frac{3}{2} \mathrm{R}$ $\therefore$ For (B): $\mathrm{P}_{\mathrm{A}}=\mathrm{I}^{2}\left(\frac{3 \mathrm{R}}{2}\right)$ $\mathrm{P}_{\mathrm{B}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{2 \mathrm{R}}+\frac{1}{\mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{2 \mathrm{R}}{3}$ $\therefore \quad \mathrm{P}_{\mathrm{B}}=\mathrm{I}^{2}\left(\frac{2 \mathrm{R}}{3}\right)$ For $(C): \mathrm{P}_{\mathrm{C}}=\mathrm{I}^{2} \mathrm{R}_{\mathrm{eq}}$ $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}$ $\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{3} \quad \mathrm{P}_{\mathrm{C}}=\mathrm{I}^{2}\left(\frac{\mathrm{R}}{3}\right)$ For (D): $\text { : } P_{P^{2}=I^{2} R_{e q}}^{R}$ Now, increasing order of power dissipation is, $\mathrm{P}_{\mathrm{C}}\lt\mathrm{P}_{\mathrm{B}}\lt\mathrm{P}_{\mathrm{A}}\lt\mathrm{P}_{\mathrm{D}}$
