QBManager

Current Electricity

151886 Find $V_{P} - V_{Q}$ in the circuit shown in figure.

1

6.68 V

2

8 V

3

4.65 V

4

7 V

Explanation:

C
$10 Ω$ and $8 Ω$ are parallel between P \& Q
$R_{PQ} = \frac{10 \times 8}{10 + 8} = \frac{80}{18} = \frac{40}{9} Ω$
$10 Ω, 18 Ω$ and $25 Ω$ are parallel between $Q & R$
$\frac{1}{R_{QR}} = \frac{1}{10} + \frac{1}{18} + \frac{1}{25} = \frac{45 + 25 + 18}{450}$
$= \frac{88}{450} = \frac{44}{225}$
$R_{QR} = \frac{225}{44} Ω$
Now, $R_{eq} = R_{PQ} + R_{QR} = \frac{40}{9} + \frac{225}{44}$
$= 9.56 Ω$
$∴ i = \frac{V}{R_{eq}} = \frac{10}{9.56} = 1.05 A$
Then, $V_{PQ} = i \times R_{PQ} = 1.05 \times \frac{40}{9}$
$= 4.65 V$

JIPMER-2018

Current Electricity

151887 $A$ and $B$ are two points on a uniform ring of resistance $19 Ω$ . The angle $∠ AOB = 45^{\circ}$ . The equivalent resistance between $A$ and $B$ is -

1

8.02 Ω

2

20.8 Ω

3

3.8 Ω

4

2.082 Ω

Explanation:

D Resistance of $360^{\circ} (R) = 19 Ω$
Resistance of $1^{\circ} = \frac{19}{360^{\circ}}$
Resistance of $45^{\circ} = \frac{19}{360^{\circ}} \times 45^{\circ}$

Then, $R_{2} = R - R_{1} = 19 - 2.375 = 16.625$
$R_{са} = \frac{R_{1} \times R_{2}}{R_{1} + R_{2}}$
$R_{еq} = \frac{2.375 \times 16.625}{2.375 + 16.625} = \frac{39.484}{19} = 2.0781 ◻ 2.082 Ω$

AP EAMCET-25.04.2018

Current Electricity

151888 Two copper wires $A$ and $B$ are measured to have same weight. If the wire $A$ is 25 times longer than wire $B$ , the ratio of their electrical resistances $(R_{A} / R_{B})$ is

1 0.04

2 25

3 625

4 225

Explanation:

C Given, $m_{A} = m_{B}, l_{A} = 25 l_{B}$
As we know that,
Resistance, $R = \frac{ρ l}{A}$
$R = \frac{ρ l \cdot l}{V}$
$R = \frac{ρ l^{2} \cdot d}{m} [∴ V = \frac{m}{d}]$
$R \propto \frac{l^{2}}{m} \dots . . (i)$
$R_{A} \propto \frac{l_{A}^{2}}{m_{A}} \dots . (ii)$
On dividing equation (i) and (ii), we get
$\frac{R_{A}}{R_{B}} = \frac{{(l_{A})}^{2} \times m_{B}}{{(l_{B})}^{2} \times m_{A}}$
$\frac{R_{A}}{R_{B}} = \frac{(25)^{2} {(l_{B})}^{2} \times m_{A}}{{(l_{B})}^{2} m_{A}} = 625$

TS EAMCET 02.05.2018

Current Electricity

151890 What will be current through the $200 Ω$ resistor in the given circuit, a long time after the switch $K$ is made on?

1 Zero

2

100 mA

3

10 mA

4

1 mA

Explanation:

C The circuit is in steady state,
So, resistance of each capacitor is infinite. It behaves like open circuit. Thus current through resistance $200 Ω$ and $400 Ω$ is same.
Now, equivalent resistance,
$R_{eq} = 200 + 400 = 600 Ω$
$∴$ Current through $200 Ω$

So, current in circuit,
$I = \frac{V}{R_{eq}}$
$I = \frac{6}{600}$
$I = 0.01 A = 10 mA$

WB JEE 2018

Current Electricity

151891 Four resistors, $100 Ω, 200 Ω, 300 Ω$ and $400 Ω$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?

1

210 Ω

2

240 Ω

3

300 Ω

4

250 Ω

Explanation:

D
For maximum equivalent resistance across the diagonal of the square, the given resistors connected as,
Resistance of ACB arm, $R_{1} = 300 + 200 = 500 Ω$
Resistance of ADB arm, $R_{2} = 400 + 100 Ω = 500 Ω$
The equivalent resistance between $A$ and $B$
$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{R_{eq}} = \frac{1}{500} + \frac{1}{500}$
$\frac{1}{R_{eq}} = \frac{2}{500}$
$R_{eq} = 250 Ω$

WB JEE 2018

Current Electricity

151886 Find $V_{P} - V_{Q}$ in the circuit shown in figure.

1

6.68 V

2

8 V

3

4.65 V

4

7 V

Explanation:

C
$10 Ω$ and $8 Ω$ are parallel between P \& Q
$R_{PQ} = \frac{10 \times 8}{10 + 8} = \frac{80}{18} = \frac{40}{9} Ω$
$10 Ω, 18 Ω$ and $25 Ω$ are parallel between $Q & R$
$\frac{1}{R_{QR}} = \frac{1}{10} + \frac{1}{18} + \frac{1}{25} = \frac{45 + 25 + 18}{450}$
$= \frac{88}{450} = \frac{44}{225}$
$R_{QR} = \frac{225}{44} Ω$
Now, $R_{eq} = R_{PQ} + R_{QR} = \frac{40}{9} + \frac{225}{44}$
$= 9.56 Ω$
$∴ i = \frac{V}{R_{eq}} = \frac{10}{9.56} = 1.05 A$
Then, $V_{PQ} = i \times R_{PQ} = 1.05 \times \frac{40}{9}$
$= 4.65 V$

JIPMER-2018

Current Electricity

151887 $A$ and $B$ are two points on a uniform ring of resistance $19 Ω$ . The angle $∠ AOB = 45^{\circ}$ . The equivalent resistance between $A$ and $B$ is -

1

8.02 Ω

2

20.8 Ω

3

3.8 Ω

4

2.082 Ω

Explanation:

D Resistance of $360^{\circ} (R) = 19 Ω$
Resistance of $1^{\circ} = \frac{19}{360^{\circ}}$
Resistance of $45^{\circ} = \frac{19}{360^{\circ}} \times 45^{\circ}$

Then, $R_{2} = R - R_{1} = 19 - 2.375 = 16.625$
$R_{са} = \frac{R_{1} \times R_{2}}{R_{1} + R_{2}}$
$R_{еq} = \frac{2.375 \times 16.625}{2.375 + 16.625} = \frac{39.484}{19} = 2.0781 ◻ 2.082 Ω$

AP EAMCET-25.04.2018

Current Electricity

151888 Two copper wires $A$ and $B$ are measured to have same weight. If the wire $A$ is 25 times longer than wire $B$ , the ratio of their electrical resistances $(R_{A} / R_{B})$ is

1 0.04

2 25

3 625

4 225

Explanation:

C Given, $m_{A} = m_{B}, l_{A} = 25 l_{B}$
As we know that,
Resistance, $R = \frac{ρ l}{A}$
$R = \frac{ρ l \cdot l}{V}$
$R = \frac{ρ l^{2} \cdot d}{m} [∴ V = \frac{m}{d}]$
$R \propto \frac{l^{2}}{m} \dots . . (i)$
$R_{A} \propto \frac{l_{A}^{2}}{m_{A}} \dots . (ii)$
On dividing equation (i) and (ii), we get
$\frac{R_{A}}{R_{B}} = \frac{{(l_{A})}^{2} \times m_{B}}{{(l_{B})}^{2} \times m_{A}}$
$\frac{R_{A}}{R_{B}} = \frac{(25)^{2} {(l_{B})}^{2} \times m_{A}}{{(l_{B})}^{2} m_{A}} = 625$

TS EAMCET 02.05.2018

Current Electricity

151890 What will be current through the $200 Ω$ resistor in the given circuit, a long time after the switch $K$ is made on?

1 Zero

2

100 mA

3

10 mA

4

1 mA

Explanation:

C The circuit is in steady state,
So, resistance of each capacitor is infinite. It behaves like open circuit. Thus current through resistance $200 Ω$ and $400 Ω$ is same.
Now, equivalent resistance,
$R_{eq} = 200 + 400 = 600 Ω$
$∴$ Current through $200 Ω$

So, current in circuit,
$I = \frac{V}{R_{eq}}$
$I = \frac{6}{600}$
$I = 0.01 A = 10 mA$

WB JEE 2018

Current Electricity

151891 Four resistors, $100 Ω, 200 Ω, 300 Ω$ and $400 Ω$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?

1

210 Ω

2

240 Ω

3

300 Ω

4

250 Ω

Explanation:

D
For maximum equivalent resistance across the diagonal of the square, the given resistors connected as,
Resistance of ACB arm, $R_{1} = 300 + 200 = 500 Ω$
Resistance of ADB arm, $R_{2} = 400 + 100 Ω = 500 Ω$
The equivalent resistance between $A$ and $B$
$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{R_{eq}} = \frac{1}{500} + \frac{1}{500}$
$\frac{1}{R_{eq}} = \frac{2}{500}$
$R_{eq} = 250 Ω$

WB JEE 2018

Current Electricity

151886 Find $V_{P} - V_{Q}$ in the circuit shown in figure.

1

6.68 V

2

8 V

3

4.65 V

4

7 V

Explanation:

C
$10 Ω$ and $8 Ω$ are parallel between P \& Q
$R_{PQ} = \frac{10 \times 8}{10 + 8} = \frac{80}{18} = \frac{40}{9} Ω$
$10 Ω, 18 Ω$ and $25 Ω$ are parallel between $Q & R$
$\frac{1}{R_{QR}} = \frac{1}{10} + \frac{1}{18} + \frac{1}{25} = \frac{45 + 25 + 18}{450}$
$= \frac{88}{450} = \frac{44}{225}$
$R_{QR} = \frac{225}{44} Ω$
Now, $R_{eq} = R_{PQ} + R_{QR} = \frac{40}{9} + \frac{225}{44}$
$= 9.56 Ω$
$∴ i = \frac{V}{R_{eq}} = \frac{10}{9.56} = 1.05 A$
Then, $V_{PQ} = i \times R_{PQ} = 1.05 \times \frac{40}{9}$
$= 4.65 V$

JIPMER-2018

Current Electricity

151887 $A$ and $B$ are two points on a uniform ring of resistance $19 Ω$ . The angle $∠ AOB = 45^{\circ}$ . The equivalent resistance between $A$ and $B$ is -

1

8.02 Ω

2

20.8 Ω

3

3.8 Ω

4

2.082 Ω

Explanation:

D Resistance of $360^{\circ} (R) = 19 Ω$
Resistance of $1^{\circ} = \frac{19}{360^{\circ}}$
Resistance of $45^{\circ} = \frac{19}{360^{\circ}} \times 45^{\circ}$

Then, $R_{2} = R - R_{1} = 19 - 2.375 = 16.625$
$R_{са} = \frac{R_{1} \times R_{2}}{R_{1} + R_{2}}$
$R_{еq} = \frac{2.375 \times 16.625}{2.375 + 16.625} = \frac{39.484}{19} = 2.0781 ◻ 2.082 Ω$

AP EAMCET-25.04.2018

Current Electricity

151888 Two copper wires $A$ and $B$ are measured to have same weight. If the wire $A$ is 25 times longer than wire $B$ , the ratio of their electrical resistances $(R_{A} / R_{B})$ is

1 0.04

2 25

3 625

4 225

Explanation:

C Given, $m_{A} = m_{B}, l_{A} = 25 l_{B}$
As we know that,
Resistance, $R = \frac{ρ l}{A}$
$R = \frac{ρ l \cdot l}{V}$
$R = \frac{ρ l^{2} \cdot d}{m} [∴ V = \frac{m}{d}]$
$R \propto \frac{l^{2}}{m} \dots . . (i)$
$R_{A} \propto \frac{l_{A}^{2}}{m_{A}} \dots . (ii)$
On dividing equation (i) and (ii), we get
$\frac{R_{A}}{R_{B}} = \frac{{(l_{A})}^{2} \times m_{B}}{{(l_{B})}^{2} \times m_{A}}$
$\frac{R_{A}}{R_{B}} = \frac{(25)^{2} {(l_{B})}^{2} \times m_{A}}{{(l_{B})}^{2} m_{A}} = 625$

TS EAMCET 02.05.2018

Current Electricity

151890 What will be current through the $200 Ω$ resistor in the given circuit, a long time after the switch $K$ is made on?

1 Zero

2

100 mA

3

10 mA

4

1 mA

Explanation:

C The circuit is in steady state,
So, resistance of each capacitor is infinite. It behaves like open circuit. Thus current through resistance $200 Ω$ and $400 Ω$ is same.
Now, equivalent resistance,
$R_{eq} = 200 + 400 = 600 Ω$
$∴$ Current through $200 Ω$

So, current in circuit,
$I = \frac{V}{R_{eq}}$
$I = \frac{6}{600}$
$I = 0.01 A = 10 mA$

WB JEE 2018

Current Electricity

151891 Four resistors, $100 Ω, 200 Ω, 300 Ω$ and $400 Ω$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?

1

210 Ω

2

240 Ω

3

300 Ω

4

250 Ω

Explanation:

D
For maximum equivalent resistance across the diagonal of the square, the given resistors connected as,
Resistance of ACB arm, $R_{1} = 300 + 200 = 500 Ω$
Resistance of ADB arm, $R_{2} = 400 + 100 Ω = 500 Ω$
The equivalent resistance between $A$ and $B$
$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{R_{eq}} = \frac{1}{500} + \frac{1}{500}$
$\frac{1}{R_{eq}} = \frac{2}{500}$
$R_{eq} = 250 Ω$

WB JEE 2018

Current Electricity

151886 Find $V_{P} - V_{Q}$ in the circuit shown in figure.

1

6.68 V

2

8 V

3

4.65 V

4

7 V

Explanation:

C
$10 Ω$ and $8 Ω$ are parallel between P \& Q
$R_{PQ} = \frac{10 \times 8}{10 + 8} = \frac{80}{18} = \frac{40}{9} Ω$
$10 Ω, 18 Ω$ and $25 Ω$ are parallel between $Q & R$
$\frac{1}{R_{QR}} = \frac{1}{10} + \frac{1}{18} + \frac{1}{25} = \frac{45 + 25 + 18}{450}$
$= \frac{88}{450} = \frac{44}{225}$
$R_{QR} = \frac{225}{44} Ω$
Now, $R_{eq} = R_{PQ} + R_{QR} = \frac{40}{9} + \frac{225}{44}$
$= 9.56 Ω$
$∴ i = \frac{V}{R_{eq}} = \frac{10}{9.56} = 1.05 A$
Then, $V_{PQ} = i \times R_{PQ} = 1.05 \times \frac{40}{9}$
$= 4.65 V$

JIPMER-2018

Current Electricity

151887 $A$ and $B$ are two points on a uniform ring of resistance $19 Ω$ . The angle $∠ AOB = 45^{\circ}$ . The equivalent resistance between $A$ and $B$ is -

1

8.02 Ω

2

20.8 Ω

3

3.8 Ω

4

2.082 Ω

Explanation:

D Resistance of $360^{\circ} (R) = 19 Ω$
Resistance of $1^{\circ} = \frac{19}{360^{\circ}}$
Resistance of $45^{\circ} = \frac{19}{360^{\circ}} \times 45^{\circ}$

Then, $R_{2} = R - R_{1} = 19 - 2.375 = 16.625$
$R_{са} = \frac{R_{1} \times R_{2}}{R_{1} + R_{2}}$
$R_{еq} = \frac{2.375 \times 16.625}{2.375 + 16.625} = \frac{39.484}{19} = 2.0781 ◻ 2.082 Ω$

AP EAMCET-25.04.2018

Current Electricity

151888 Two copper wires $A$ and $B$ are measured to have same weight. If the wire $A$ is 25 times longer than wire $B$ , the ratio of their electrical resistances $(R_{A} / R_{B})$ is

1 0.04

2 25

3 625

4 225

Explanation:

C Given, $m_{A} = m_{B}, l_{A} = 25 l_{B}$
As we know that,
Resistance, $R = \frac{ρ l}{A}$
$R = \frac{ρ l \cdot l}{V}$
$R = \frac{ρ l^{2} \cdot d}{m} [∴ V = \frac{m}{d}]$
$R \propto \frac{l^{2}}{m} \dots . . (i)$
$R_{A} \propto \frac{l_{A}^{2}}{m_{A}} \dots . (ii)$
On dividing equation (i) and (ii), we get
$\frac{R_{A}}{R_{B}} = \frac{{(l_{A})}^{2} \times m_{B}}{{(l_{B})}^{2} \times m_{A}}$
$\frac{R_{A}}{R_{B}} = \frac{(25)^{2} {(l_{B})}^{2} \times m_{A}}{{(l_{B})}^{2} m_{A}} = 625$

TS EAMCET 02.05.2018

Current Electricity

151890 What will be current through the $200 Ω$ resistor in the given circuit, a long time after the switch $K$ is made on?

1 Zero

2

100 mA

3

10 mA

4

1 mA

Explanation:

C The circuit is in steady state,
So, resistance of each capacitor is infinite. It behaves like open circuit. Thus current through resistance $200 Ω$ and $400 Ω$ is same.
Now, equivalent resistance,
$R_{eq} = 200 + 400 = 600 Ω$
$∴$ Current through $200 Ω$

So, current in circuit,
$I = \frac{V}{R_{eq}}$
$I = \frac{6}{600}$
$I = 0.01 A = 10 mA$

WB JEE 2018

Current Electricity

151891 Four resistors, $100 Ω, 200 Ω, 300 Ω$ and $400 Ω$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?

1

210 Ω

2

240 Ω

3

300 Ω

4

250 Ω

Explanation:

D
For maximum equivalent resistance across the diagonal of the square, the given resistors connected as,
Resistance of ACB arm, $R_{1} = 300 + 200 = 500 Ω$
Resistance of ADB arm, $R_{2} = 400 + 100 Ω = 500 Ω$
The equivalent resistance between $A$ and $B$
$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{R_{eq}} = \frac{1}{500} + \frac{1}{500}$
$\frac{1}{R_{eq}} = \frac{2}{500}$
$R_{eq} = 250 Ω$

WB JEE 2018

Current Electricity

151886 Find $V_{P} - V_{Q}$ in the circuit shown in figure.

1

6.68 V

2

8 V

3

4.65 V

4

7 V

Explanation:

C
$10 Ω$ and $8 Ω$ are parallel between P \& Q
$R_{PQ} = \frac{10 \times 8}{10 + 8} = \frac{80}{18} = \frac{40}{9} Ω$
$10 Ω, 18 Ω$ and $25 Ω$ are parallel between $Q & R$
$\frac{1}{R_{QR}} = \frac{1}{10} + \frac{1}{18} + \frac{1}{25} = \frac{45 + 25 + 18}{450}$
$= \frac{88}{450} = \frac{44}{225}$
$R_{QR} = \frac{225}{44} Ω$
Now, $R_{eq} = R_{PQ} + R_{QR} = \frac{40}{9} + \frac{225}{44}$
$= 9.56 Ω$
$∴ i = \frac{V}{R_{eq}} = \frac{10}{9.56} = 1.05 A$
Then, $V_{PQ} = i \times R_{PQ} = 1.05 \times \frac{40}{9}$
$= 4.65 V$

JIPMER-2018

Current Electricity

151887 $A$ and $B$ are two points on a uniform ring of resistance $19 Ω$ . The angle $∠ AOB = 45^{\circ}$ . The equivalent resistance between $A$ and $B$ is -

1

8.02 Ω

2

20.8 Ω

3

3.8 Ω

4

2.082 Ω

Explanation:

D Resistance of $360^{\circ} (R) = 19 Ω$
Resistance of $1^{\circ} = \frac{19}{360^{\circ}}$
Resistance of $45^{\circ} = \frac{19}{360^{\circ}} \times 45^{\circ}$

Then, $R_{2} = R - R_{1} = 19 - 2.375 = 16.625$
$R_{са} = \frac{R_{1} \times R_{2}}{R_{1} + R_{2}}$
$R_{еq} = \frac{2.375 \times 16.625}{2.375 + 16.625} = \frac{39.484}{19} = 2.0781 ◻ 2.082 Ω$

AP EAMCET-25.04.2018

Current Electricity

151888 Two copper wires $A$ and $B$ are measured to have same weight. If the wire $A$ is 25 times longer than wire $B$ , the ratio of their electrical resistances $(R_{A} / R_{B})$ is

1 0.04

2 25

3 625

4 225

Explanation:

C Given, $m_{A} = m_{B}, l_{A} = 25 l_{B}$
As we know that,
Resistance, $R = \frac{ρ l}{A}$
$R = \frac{ρ l \cdot l}{V}$
$R = \frac{ρ l^{2} \cdot d}{m} [∴ V = \frac{m}{d}]$
$R \propto \frac{l^{2}}{m} \dots . . (i)$
$R_{A} \propto \frac{l_{A}^{2}}{m_{A}} \dots . (ii)$
On dividing equation (i) and (ii), we get
$\frac{R_{A}}{R_{B}} = \frac{{(l_{A})}^{2} \times m_{B}}{{(l_{B})}^{2} \times m_{A}}$
$\frac{R_{A}}{R_{B}} = \frac{(25)^{2} {(l_{B})}^{2} \times m_{A}}{{(l_{B})}^{2} m_{A}} = 625$

TS EAMCET 02.05.2018

Current Electricity

151890 What will be current through the $200 Ω$ resistor in the given circuit, a long time after the switch $K$ is made on?

1 Zero

2

100 mA

3

10 mA

4

1 mA

Explanation:

C The circuit is in steady state,
So, resistance of each capacitor is infinite. It behaves like open circuit. Thus current through resistance $200 Ω$ and $400 Ω$ is same.
Now, equivalent resistance,
$R_{eq} = 200 + 400 = 600 Ω$
$∴$ Current through $200 Ω$

So, current in circuit,
$I = \frac{V}{R_{eq}}$
$I = \frac{6}{600}$
$I = 0.01 A = 10 mA$

WB JEE 2018

Current Electricity

151891 Four resistors, $100 Ω, 200 Ω, 300 Ω$ and $400 Ω$ are connected to form four sides of a square. The resistors can be connected in any order. What is the maximum possible equivalent resistance across the diagonal of the square?

1

210 Ω

2

240 Ω

3

300 Ω

4

250 Ω

Explanation:

D
For maximum equivalent resistance across the diagonal of the square, the given resistors connected as,
Resistance of ACB arm, $R_{1} = 300 + 200 = 500 Ω$
Resistance of ADB arm, $R_{2} = 400 + 100 Ω = 500 Ω$
The equivalent resistance between $A$ and $B$
$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{R_{eq}} = \frac{1}{500} + \frac{1}{500}$
$\frac{1}{R_{eq}} = \frac{2}{500}$
$R_{eq} = 250 Ω$

WB JEE 2018