As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of . The currents through the resistors and are and respectively. The values of and are [2023]

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Q.
As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of $3 Ω$ . The currents through the resistors $R_{4}$ and $R_{5}$ are $I_{4}$ and $I_{5}$ respectively. The values of $I_{4}$ and $I_{5}$ are [2023]

1 $I_{4} = \frac{8}{5} A$ and $I_{5} = \frac{2}{5} A$

2 $I_{4} = \frac{6}{5} A$ and $I_{5} = \frac{24}{5} A$

3 $I_{4} = \frac{2}{5} A$ and $I_{5} = \frac{8}{5} A$

4 $I_{4} = \frac{24}{5} A$ and $I_{5} = \frac{6}{5} A$

Ans.
(3)

$Equivalent resistance of circuit$

$R_{eq} = 3 + 1 + 2 + 4 + 2 = 12 Ω$

$Current through battery i = \frac{24}{12} = 2 A$

$I_{4} = \frac{R_{5}}{R_{4} + R_{5}} \times 2 = \frac{5}{20 + 5} \times 2 = \frac{2}{5} A$

$I_{5} = 2 - \frac{2}{5} = \frac{8}{5} A$

Solution

Q.
As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of $3 Ω$ . The currents through the resistors $R_{4}$ and $R_{5}$ are $I_{4}$ and $I_{5}$ respectively. The values of $I_{4}$ and $I_{5}$ are [2023]

1 $I_{4} = \frac{8}{5} A$ and $I_{5} = \frac{2}{5} A$

2 $I_{4} = \frac{6}{5} A$ and $I_{5} = \frac{24}{5} A$

3 $I_{4} = \frac{2}{5} A$ and $I_{5} = \frac{8}{5} A$

4 $I_{4} = \frac{24}{5} A$ and $I_{5} = \frac{6}{5} A$

Ans.
(3)

$Equivalent resistance of circuit$

$R_{eq} = 3 + 1 + 2 + 4 + 2 = 12 Ω$

$Current through battery i = \frac{24}{12} = 2 A$

$I_{4} = \frac{R_{5}}{R_{4} + R_{5}} \times 2 = \frac{5}{20 + 5} \times 2 = \frac{2}{5} A$

$I_{5} = 2 - \frac{2}{5} = \frac{8}{5} A$

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Solution

Q. As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of 3 Ω. The currents through the resistors R4 and R5 are I4 and I5 respectively. The values of I4 and I5 are [2023]

1 I4=85A and I5=25A

2 I4=65A and I5=245A

3 I4=25A and I5=85A

4 I4=245A and I5=65A

Ans. (3) Equivalent resistance of circuit Req=3+1+2+4+2=12 Ω Current through battery i=2412=2 A I4=R5R4+R5×2=520+5×2=25 A I5=2-25=85 A

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Q.
As shown in the figure, a network of resistors is connected to a battery of 24 V with an internal resistance of $3 Ω$ . The currents through the resistors $R_{4}$ and $R_{5}$ are $I_{4}$ and $I_{5}$ respectively. The values of $I_{4}$ and $I_{5}$ are [2023]

1 $I_{4} = \frac{8}{5} A$ and $I_{5} = \frac{2}{5} A$

2 $I_{4} = \frac{6}{5} A$ and $I_{5} = \frac{24}{5} A$

3 $I_{4} = \frac{2}{5} A$ and $I_{5} = \frac{8}{5} A$

4 $I_{4} = \frac{24}{5} A$ and $I_{5} = \frac{6}{5} A$

Ans.
(3)

$Equivalent resistance of circuit$

$R_{eq} = 3 + 1 + 2 + 4 + 2 = 12 Ω$

$Current through battery i = \frac{24}{12} = 2 A$

$I_{4} = \frac{R_{5}}{R_{4} + R_{5}} \times 2 = \frac{5}{20 + 5} \times 2 = \frac{2}{5} A$

$I_{5} = 2 - \frac{2}{5} = \frac{8}{5} A$