QBManager

Current Electricity

152814 With resistance $P$ and $Q$ in the left and the right gap respectively of a meter bridge, the null point divides the wire in the ratio $3 : 4$ . When $P$ and $Q$ are increased by $20 Ω$ each, the null point divides the wire in the ratio $5 : 6$ . Then the values of $P$ . $Q$ respectively are .............

1

30 Ω, 40 Ω

2

20 Ω, 40 Ω

3

30 Ω, 80 Ω

4

20 Ω, 20 Ω

Explanation:

A For balanced bridge $\frac{x}{R} = \frac{L_{x}}{L_{r}}$
$\frac{x}{R} = \frac{3}{4}$
$\frac{P}{Q} = \frac{3}{4}$
$3 Q = 4 P$
$3 Q - 4 P = 0$
$Now, \frac{P + 20}{Q + 20} = \frac{5}{6}$
$6 (P + 20) = 5 (Q + 20)$
$6 P + 120 = 5 Q + 120$
$5 Q - 6 P = 20$
Solving eq $^{n}$ (i) \& eq $^{n}$ (ii) we have.
Multiplying 3 in eq (i) and by 2 in eq $^{n}$ (ii)
$9 Q - 12 P = 0$
$10 Q - 12 P = 40$
$Q = 40 Ω$
Putting the value of $Q$ in eq $^{n}$ (i) we have,
$3 \times 40 - 4 P = 0$
$120 - 4 P = 0$
$4 P = 120$
$P = 30 Ω$

AP EAMCET-23.09.2020

Current Electricity

152816 A voltmeter has a resistance ' $G$ ' and range ' $V$ '. What resistance should be connected in series to it. in order to increase its range to $nV$ ?

1

nG

2

(n + 1) G

3

(n - 1) G

4

\frac{G}{n}

Explanation:

C A high resistance is connected in series with coil of the galvanometer to convert it to voltmeter.
Resistance connected in Series is given by
Maximum current through the galvanometer

$I_{g} = \frac{V}{G}$
$V_{1} = (R_{S} + G) I_{g} = n \cdot V$
$(R_{S} + G) \frac{V}{G} = n V$
$R_{S} = (n - 1) \cdot G$

AP EAMCET-24.09.2020

Current Electricity

152817 An ammeter shows full deflection when a current of $1 A$ is passing through it, $A$ shunt is connected such that it shows full deflection for a current of $10 A$ . The ratio of resistance of the shunt and the ammeter is

1

9 : 1

2

1 : 9

3

11 : 1

4

1 : 11

Explanation:

B An Ammeter gives full scale deflection with a current of 1 amp $r_{1} = \frac{V}{1} = V$
Ammeter of range $r_{2} = \frac{V}{10} = 0.1 V$
Ratio of resistance of the shunt and ammeter
$\frac{r_{1}}{r_{2}} = \frac{0.1 V}{V - 0.1 V}$
$\frac{r_{1}}{r_{2}} = \frac{0.1 V}{0.9 V} = \frac{1}{9}$
$r_{1} : r_{2} = 1 : 9$

Tripura-2020

Current Electricity

152818 In the circuit shown in figure, the ammeter $A$ reads zero. If the batteries box has negligible internal resistances, the value of $R$ is

1

5 Ω

2

10 Ω

3

15 Ω

4

20 Ω

Explanation:

D
Current due to $E_{1}$ is $I$
$I = \frac{E_{1}}{R + 60}$
$= \frac{4}{R + 60}$
Ammeter reads zero as there is no current due to $E_{2}$ .
$∴ IR = E_{2} = 1 V$
$∴ I = \frac{1}{R}$
From equation (i) and (ii)-
$\frac{4}{R + 60} = \frac{1}{R}$
$R + 60 = 4 R$
$3 R = 60$
$R = 20 Ω$

Assam CEE-2020

Current Electricity

152820 A galvanometer can be converted to a voltmeter of full scale deflection $V_{0}$ by connecting a series resistance $R_{1}$ and can be converted to an ammeter of full scale deflection $I_{0}$ by connecting a shunt resistance $R_{2}$ . What is the current flowing through the galvanometer at its full scale deflection?

1

\frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}

2

\frac{V_{0} + I_{0} R_{2}}{R_{1} + R_{2}}

3

\frac{V_{0} - I_{0} R_{1}}{R_{2} - R_{1}}

4

\frac{V_{0} + I_{0} R_{1}}{R_{1} + R_{2}}

Explanation:

A For voltmeter
$V = (G + R_{1}) I_{g}$
Where, $I_{g}$ is current through the galvanometer for Ammeter shunt $S = R_{2}$
$R_{2} = \frac{G}{\frac{I_{0}}{I_{G}} - 1} = \frac{{GI}_{G}}{I_{0} - I_{g}}$
On solving equation (i) and (ii)
$V_{0} = {GI}_{g} + R_{1} I_{g}$
$V_{0} = R_{2} (I_{0} - I_{g}) + R_{1} I_{g}$
$I_{g} = \frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}$

WB JEE 2020

Current Electricity

152814 With resistance $P$ and $Q$ in the left and the right gap respectively of a meter bridge, the null point divides the wire in the ratio $3 : 4$ . When $P$ and $Q$ are increased by $20 Ω$ each, the null point divides the wire in the ratio $5 : 6$ . Then the values of $P$ . $Q$ respectively are .............

1

30 Ω, 40 Ω

2

20 Ω, 40 Ω

3

30 Ω, 80 Ω

4

20 Ω, 20 Ω

Explanation:

A For balanced bridge $\frac{x}{R} = \frac{L_{x}}{L_{r}}$
$\frac{x}{R} = \frac{3}{4}$
$\frac{P}{Q} = \frac{3}{4}$
$3 Q = 4 P$
$3 Q - 4 P = 0$
$Now, \frac{P + 20}{Q + 20} = \frac{5}{6}$
$6 (P + 20) = 5 (Q + 20)$
$6 P + 120 = 5 Q + 120$
$5 Q - 6 P = 20$
Solving eq $^{n}$ (i) \& eq $^{n}$ (ii) we have.
Multiplying 3 in eq (i) and by 2 in eq $^{n}$ (ii)
$9 Q - 12 P = 0$
$10 Q - 12 P = 40$
$Q = 40 Ω$
Putting the value of $Q$ in eq $^{n}$ (i) we have,
$3 \times 40 - 4 P = 0$
$120 - 4 P = 0$
$4 P = 120$
$P = 30 Ω$

AP EAMCET-23.09.2020

Current Electricity

152816 A voltmeter has a resistance ' $G$ ' and range ' $V$ '. What resistance should be connected in series to it. in order to increase its range to $nV$ ?

1

nG

2

(n + 1) G

3

(n - 1) G

4

\frac{G}{n}

Explanation:

C A high resistance is connected in series with coil of the galvanometer to convert it to voltmeter.
Resistance connected in Series is given by
Maximum current through the galvanometer

$I_{g} = \frac{V}{G}$
$V_{1} = (R_{S} + G) I_{g} = n \cdot V$
$(R_{S} + G) \frac{V}{G} = n V$
$R_{S} = (n - 1) \cdot G$

AP EAMCET-24.09.2020

Current Electricity

152817 An ammeter shows full deflection when a current of $1 A$ is passing through it, $A$ shunt is connected such that it shows full deflection for a current of $10 A$ . The ratio of resistance of the shunt and the ammeter is

1

9 : 1

2

1 : 9

3

11 : 1

4

1 : 11

Explanation:

B An Ammeter gives full scale deflection with a current of 1 amp $r_{1} = \frac{V}{1} = V$
Ammeter of range $r_{2} = \frac{V}{10} = 0.1 V$
Ratio of resistance of the shunt and ammeter
$\frac{r_{1}}{r_{2}} = \frac{0.1 V}{V - 0.1 V}$
$\frac{r_{1}}{r_{2}} = \frac{0.1 V}{0.9 V} = \frac{1}{9}$
$r_{1} : r_{2} = 1 : 9$

Tripura-2020

Current Electricity

152818 In the circuit shown in figure, the ammeter $A$ reads zero. If the batteries box has negligible internal resistances, the value of $R$ is

1

5 Ω

2

10 Ω

3

15 Ω

4

20 Ω

Explanation:

D
Current due to $E_{1}$ is $I$
$I = \frac{E_{1}}{R + 60}$
$= \frac{4}{R + 60}$
Ammeter reads zero as there is no current due to $E_{2}$ .
$∴ IR = E_{2} = 1 V$
$∴ I = \frac{1}{R}$
From equation (i) and (ii)-
$\frac{4}{R + 60} = \frac{1}{R}$
$R + 60 = 4 R$
$3 R = 60$
$R = 20 Ω$

Assam CEE-2020

Current Electricity

152820 A galvanometer can be converted to a voltmeter of full scale deflection $V_{0}$ by connecting a series resistance $R_{1}$ and can be converted to an ammeter of full scale deflection $I_{0}$ by connecting a shunt resistance $R_{2}$ . What is the current flowing through the galvanometer at its full scale deflection?

1

\frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}

2

\frac{V_{0} + I_{0} R_{2}}{R_{1} + R_{2}}

3

\frac{V_{0} - I_{0} R_{1}}{R_{2} - R_{1}}

4

\frac{V_{0} + I_{0} R_{1}}{R_{1} + R_{2}}

Explanation:

A For voltmeter
$V = (G + R_{1}) I_{g}$
Where, $I_{g}$ is current through the galvanometer for Ammeter shunt $S = R_{2}$
$R_{2} = \frac{G}{\frac{I_{0}}{I_{G}} - 1} = \frac{{GI}_{G}}{I_{0} - I_{g}}$
On solving equation (i) and (ii)
$V_{0} = {GI}_{g} + R_{1} I_{g}$
$V_{0} = R_{2} (I_{0} - I_{g}) + R_{1} I_{g}$
$I_{g} = \frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}$

WB JEE 2020

Current Electricity

152814 With resistance $P$ and $Q$ in the left and the right gap respectively of a meter bridge, the null point divides the wire in the ratio $3 : 4$ . When $P$ and $Q$ are increased by $20 Ω$ each, the null point divides the wire in the ratio $5 : 6$ . Then the values of $P$ . $Q$ respectively are .............

1

30 Ω, 40 Ω

2

20 Ω, 40 Ω

3

30 Ω, 80 Ω

4

20 Ω, 20 Ω

Explanation:

A For balanced bridge $\frac{x}{R} = \frac{L_{x}}{L_{r}}$
$\frac{x}{R} = \frac{3}{4}$
$\frac{P}{Q} = \frac{3}{4}$
$3 Q = 4 P$
$3 Q - 4 P = 0$
$Now, \frac{P + 20}{Q + 20} = \frac{5}{6}$
$6 (P + 20) = 5 (Q + 20)$
$6 P + 120 = 5 Q + 120$
$5 Q - 6 P = 20$
Solving eq $^{n}$ (i) \& eq $^{n}$ (ii) we have.
Multiplying 3 in eq (i) and by 2 in eq $^{n}$ (ii)
$9 Q - 12 P = 0$
$10 Q - 12 P = 40$
$Q = 40 Ω$
Putting the value of $Q$ in eq $^{n}$ (i) we have,
$3 \times 40 - 4 P = 0$
$120 - 4 P = 0$
$4 P = 120$
$P = 30 Ω$

AP EAMCET-23.09.2020

Current Electricity

152816 A voltmeter has a resistance ' $G$ ' and range ' $V$ '. What resistance should be connected in series to it. in order to increase its range to $nV$ ?

1

nG

2

(n + 1) G

3

(n - 1) G

4

\frac{G}{n}

Explanation:

C A high resistance is connected in series with coil of the galvanometer to convert it to voltmeter.
Resistance connected in Series is given by
Maximum current through the galvanometer

$I_{g} = \frac{V}{G}$
$V_{1} = (R_{S} + G) I_{g} = n \cdot V$
$(R_{S} + G) \frac{V}{G} = n V$
$R_{S} = (n - 1) \cdot G$

AP EAMCET-24.09.2020

Current Electricity

152817 An ammeter shows full deflection when a current of $1 A$ is passing through it, $A$ shunt is connected such that it shows full deflection for a current of $10 A$ . The ratio of resistance of the shunt and the ammeter is

1

9 : 1

2

1 : 9

3

11 : 1

4

1 : 11

Explanation:

B An Ammeter gives full scale deflection with a current of 1 amp $r_{1} = \frac{V}{1} = V$
Ammeter of range $r_{2} = \frac{V}{10} = 0.1 V$
Ratio of resistance of the shunt and ammeter
$\frac{r_{1}}{r_{2}} = \frac{0.1 V}{V - 0.1 V}$
$\frac{r_{1}}{r_{2}} = \frac{0.1 V}{0.9 V} = \frac{1}{9}$
$r_{1} : r_{2} = 1 : 9$

Tripura-2020

Current Electricity

152818 In the circuit shown in figure, the ammeter $A$ reads zero. If the batteries box has negligible internal resistances, the value of $R$ is

1

5 Ω

2

10 Ω

3

15 Ω

4

20 Ω

Explanation:

D
Current due to $E_{1}$ is $I$
$I = \frac{E_{1}}{R + 60}$
$= \frac{4}{R + 60}$
Ammeter reads zero as there is no current due to $E_{2}$ .
$∴ IR = E_{2} = 1 V$
$∴ I = \frac{1}{R}$
From equation (i) and (ii)-
$\frac{4}{R + 60} = \frac{1}{R}$
$R + 60 = 4 R$
$3 R = 60$
$R = 20 Ω$

Assam CEE-2020

Current Electricity

152820 A galvanometer can be converted to a voltmeter of full scale deflection $V_{0}$ by connecting a series resistance $R_{1}$ and can be converted to an ammeter of full scale deflection $I_{0}$ by connecting a shunt resistance $R_{2}$ . What is the current flowing through the galvanometer at its full scale deflection?

1

\frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}

2

\frac{V_{0} + I_{0} R_{2}}{R_{1} + R_{2}}

3

\frac{V_{0} - I_{0} R_{1}}{R_{2} - R_{1}}

4

\frac{V_{0} + I_{0} R_{1}}{R_{1} + R_{2}}

Explanation:

A For voltmeter
$V = (G + R_{1}) I_{g}$
Where, $I_{g}$ is current through the galvanometer for Ammeter shunt $S = R_{2}$
$R_{2} = \frac{G}{\frac{I_{0}}{I_{G}} - 1} = \frac{{GI}_{G}}{I_{0} - I_{g}}$
On solving equation (i) and (ii)
$V_{0} = {GI}_{g} + R_{1} I_{g}$
$V_{0} = R_{2} (I_{0} - I_{g}) + R_{1} I_{g}$
$I_{g} = \frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}$

WB JEE 2020

Current Electricity

152814 With resistance $P$ and $Q$ in the left and the right gap respectively of a meter bridge, the null point divides the wire in the ratio $3 : 4$ . When $P$ and $Q$ are increased by $20 Ω$ each, the null point divides the wire in the ratio $5 : 6$ . Then the values of $P$ . $Q$ respectively are .............

1

30 Ω, 40 Ω

2

20 Ω, 40 Ω

3

30 Ω, 80 Ω

4

20 Ω, 20 Ω

Explanation:

A For balanced bridge $\frac{x}{R} = \frac{L_{x}}{L_{r}}$
$\frac{x}{R} = \frac{3}{4}$
$\frac{P}{Q} = \frac{3}{4}$
$3 Q = 4 P$
$3 Q - 4 P = 0$
$Now, \frac{P + 20}{Q + 20} = \frac{5}{6}$
$6 (P + 20) = 5 (Q + 20)$
$6 P + 120 = 5 Q + 120$
$5 Q - 6 P = 20$
Solving eq $^{n}$ (i) \& eq $^{n}$ (ii) we have.
Multiplying 3 in eq (i) and by 2 in eq $^{n}$ (ii)
$9 Q - 12 P = 0$
$10 Q - 12 P = 40$
$Q = 40 Ω$
Putting the value of $Q$ in eq $^{n}$ (i) we have,
$3 \times 40 - 4 P = 0$
$120 - 4 P = 0$
$4 P = 120$
$P = 30 Ω$

AP EAMCET-23.09.2020

Current Electricity

152816 A voltmeter has a resistance ' $G$ ' and range ' $V$ '. What resistance should be connected in series to it. in order to increase its range to $nV$ ?

1

nG

2

(n + 1) G

3

(n - 1) G

4

\frac{G}{n}

Explanation:

C A high resistance is connected in series with coil of the galvanometer to convert it to voltmeter.
Resistance connected in Series is given by
Maximum current through the galvanometer

$I_{g} = \frac{V}{G}$
$V_{1} = (R_{S} + G) I_{g} = n \cdot V$
$(R_{S} + G) \frac{V}{G} = n V$
$R_{S} = (n - 1) \cdot G$

AP EAMCET-24.09.2020

Current Electricity

152817 An ammeter shows full deflection when a current of $1 A$ is passing through it, $A$ shunt is connected such that it shows full deflection for a current of $10 A$ . The ratio of resistance of the shunt and the ammeter is

1

9 : 1

2

1 : 9

3

11 : 1

4

1 : 11

Explanation:

B An Ammeter gives full scale deflection with a current of 1 amp $r_{1} = \frac{V}{1} = V$
Ammeter of range $r_{2} = \frac{V}{10} = 0.1 V$
Ratio of resistance of the shunt and ammeter
$\frac{r_{1}}{r_{2}} = \frac{0.1 V}{V - 0.1 V}$
$\frac{r_{1}}{r_{2}} = \frac{0.1 V}{0.9 V} = \frac{1}{9}$
$r_{1} : r_{2} = 1 : 9$

Tripura-2020

Current Electricity

152818 In the circuit shown in figure, the ammeter $A$ reads zero. If the batteries box has negligible internal resistances, the value of $R$ is

1

5 Ω

2

10 Ω

3

15 Ω

4

20 Ω

Explanation:

D
Current due to $E_{1}$ is $I$
$I = \frac{E_{1}}{R + 60}$
$= \frac{4}{R + 60}$
Ammeter reads zero as there is no current due to $E_{2}$ .
$∴ IR = E_{2} = 1 V$
$∴ I = \frac{1}{R}$
From equation (i) and (ii)-
$\frac{4}{R + 60} = \frac{1}{R}$
$R + 60 = 4 R$
$3 R = 60$
$R = 20 Ω$

Assam CEE-2020

Current Electricity

152820 A galvanometer can be converted to a voltmeter of full scale deflection $V_{0}$ by connecting a series resistance $R_{1}$ and can be converted to an ammeter of full scale deflection $I_{0}$ by connecting a shunt resistance $R_{2}$ . What is the current flowing through the galvanometer at its full scale deflection?

1

\frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}

2

\frac{V_{0} + I_{0} R_{2}}{R_{1} + R_{2}}

3

\frac{V_{0} - I_{0} R_{1}}{R_{2} - R_{1}}

4

\frac{V_{0} + I_{0} R_{1}}{R_{1} + R_{2}}

Explanation:

A For voltmeter
$V = (G + R_{1}) I_{g}$
Where, $I_{g}$ is current through the galvanometer for Ammeter shunt $S = R_{2}$
$R_{2} = \frac{G}{\frac{I_{0}}{I_{G}} - 1} = \frac{{GI}_{G}}{I_{0} - I_{g}}$
On solving equation (i) and (ii)
$V_{0} = {GI}_{g} + R_{1} I_{g}$
$V_{0} = R_{2} (I_{0} - I_{g}) + R_{1} I_{g}$
$I_{g} = \frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}$

WB JEE 2020

Current Electricity

152814 With resistance $P$ and $Q$ in the left and the right gap respectively of a meter bridge, the null point divides the wire in the ratio $3 : 4$ . When $P$ and $Q$ are increased by $20 Ω$ each, the null point divides the wire in the ratio $5 : 6$ . Then the values of $P$ . $Q$ respectively are .............

1

30 Ω, 40 Ω

2

20 Ω, 40 Ω

3

30 Ω, 80 Ω

4

20 Ω, 20 Ω

Explanation:

A For balanced bridge $\frac{x}{R} = \frac{L_{x}}{L_{r}}$
$\frac{x}{R} = \frac{3}{4}$
$\frac{P}{Q} = \frac{3}{4}$
$3 Q = 4 P$
$3 Q - 4 P = 0$
$Now, \frac{P + 20}{Q + 20} = \frac{5}{6}$
$6 (P + 20) = 5 (Q + 20)$
$6 P + 120 = 5 Q + 120$
$5 Q - 6 P = 20$
Solving eq $^{n}$ (i) \& eq $^{n}$ (ii) we have.
Multiplying 3 in eq (i) and by 2 in eq $^{n}$ (ii)
$9 Q - 12 P = 0$
$10 Q - 12 P = 40$
$Q = 40 Ω$
Putting the value of $Q$ in eq $^{n}$ (i) we have,
$3 \times 40 - 4 P = 0$
$120 - 4 P = 0$
$4 P = 120$
$P = 30 Ω$

AP EAMCET-23.09.2020

Current Electricity

152816 A voltmeter has a resistance ' $G$ ' and range ' $V$ '. What resistance should be connected in series to it. in order to increase its range to $nV$ ?

1

nG

2

(n + 1) G

3

(n - 1) G

4

\frac{G}{n}

Explanation:

C A high resistance is connected in series with coil of the galvanometer to convert it to voltmeter.
Resistance connected in Series is given by
Maximum current through the galvanometer

$I_{g} = \frac{V}{G}$
$V_{1} = (R_{S} + G) I_{g} = n \cdot V$
$(R_{S} + G) \frac{V}{G} = n V$
$R_{S} = (n - 1) \cdot G$

AP EAMCET-24.09.2020

Current Electricity

152817 An ammeter shows full deflection when a current of $1 A$ is passing through it, $A$ shunt is connected such that it shows full deflection for a current of $10 A$ . The ratio of resistance of the shunt and the ammeter is

1

9 : 1

2

1 : 9

3

11 : 1

4

1 : 11

Explanation:

B An Ammeter gives full scale deflection with a current of 1 amp $r_{1} = \frac{V}{1} = V$
Ammeter of range $r_{2} = \frac{V}{10} = 0.1 V$
Ratio of resistance of the shunt and ammeter
$\frac{r_{1}}{r_{2}} = \frac{0.1 V}{V - 0.1 V}$
$\frac{r_{1}}{r_{2}} = \frac{0.1 V}{0.9 V} = \frac{1}{9}$
$r_{1} : r_{2} = 1 : 9$

Tripura-2020

Current Electricity

152818 In the circuit shown in figure, the ammeter $A$ reads zero. If the batteries box has negligible internal resistances, the value of $R$ is

1

5 Ω

2

10 Ω

3

15 Ω

4

20 Ω

Explanation:

D
Current due to $E_{1}$ is $I$
$I = \frac{E_{1}}{R + 60}$
$= \frac{4}{R + 60}$
Ammeter reads zero as there is no current due to $E_{2}$ .
$∴ IR = E_{2} = 1 V$
$∴ I = \frac{1}{R}$
From equation (i) and (ii)-
$\frac{4}{R + 60} = \frac{1}{R}$
$R + 60 = 4 R$
$3 R = 60$
$R = 20 Ω$

Assam CEE-2020

Current Electricity

152820 A galvanometer can be converted to a voltmeter of full scale deflection $V_{0}$ by connecting a series resistance $R_{1}$ and can be converted to an ammeter of full scale deflection $I_{0}$ by connecting a shunt resistance $R_{2}$ . What is the current flowing through the galvanometer at its full scale deflection?

1

\frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}

2

\frac{V_{0} + I_{0} R_{2}}{R_{1} + R_{2}}

3

\frac{V_{0} - I_{0} R_{1}}{R_{2} - R_{1}}

4

\frac{V_{0} + I_{0} R_{1}}{R_{1} + R_{2}}

Explanation:

A For voltmeter
$V = (G + R_{1}) I_{g}$
Where, $I_{g}$ is current through the galvanometer for Ammeter shunt $S = R_{2}$
$R_{2} = \frac{G}{\frac{I_{0}}{I_{G}} - 1} = \frac{{GI}_{G}}{I_{0} - I_{g}}$
On solving equation (i) and (ii)
$V_{0} = {GI}_{g} + R_{1} I_{g}$
$V_{0} = R_{2} (I_{0} - I_{g}) + R_{1} I_{g}$
$I_{g} = \frac{V_{0} - I_{0} R_{2}}{R_{1} - R_{2}}$

WB JEE 2020

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: