SATHEE: Current Electricity 3 Question 14

Current Electricity 3 Question 14

13. For the circuit shown in the figure

(2009)

Show Answer

Solution:

$R_{total} = 2 + \frac{6 \times 1.5}{6 + 1.5} = 3.2 k Ω$

(a) $I = \frac{24 V}{3.2 k Ω} = 7.5 m A = I_{R_{1}}$

$I_{R_{2}} = \frac{R_{L}}{R_{L} + R_{2}} I$

$I = \frac{1.5}{7.5} \times 7.5 = 1.5 m A$

$I_{R_{L}} = 6 m A$

(b) $V_{R_{L}} = (I_{R_{L}}) (R_{L}) = 9 V$

(c) $\frac{P_{R_{1}}}{P_{R_{2}}} = \frac{(I_{R_{1}}^{2}) R_{1}}{(I_{R_{2}}^{2}) R_{2}} = \frac{(7.5)^{2} (2)}{(1.5)^{2} (6)} = \frac{25}{3}$

(d) When $R_{1}$ and $R_{2}$ are inter changed, then

$\frac{R_{2} R_{L}}{R_{2} + R_{L}} = \frac{2 \times 1.5}{3.5} = \frac{6}{7} k Ω$

Now potential difference across $R_{L}$ will be

$V_{L} = 24 \frac{6 / 7}{6 + 6 / 7} = 3 V$

Earlier it was $9 V$

Since, $P = \frac{V^{2}}{R}$ or $P \propto V^{2}$

In new situation potential difference has been decreased three times. Therefore, power dissipated will decrease by a factor of 9.

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Current Electricity 3 Question 14

13. For the circuit shown in the figure

Solution:

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