For the circuit shown in the figure [2009]

Question

For the circuit shown in the figure [2009]

Smart Achievers · Accepted Answer

(1, 4)

$R_{p} = \frac{R_{2} \times R_{L}}{R_{2} + R_{L}} = \frac{6 \times 1.5}{6 + 1.5} = \frac{9}{7.5} k Ω$

$∴$ $R_{total} = R_{1} + R_{p} = 2 + \frac{9}{7.5} = 3.2 k Ω$

$(1)$ $I = \frac{V}{R} = \frac{24}{3.2} mA = 7.5 mA$

$I = I_{R_{1}}$

$I_{R_{2}} = (\frac{R_{L}}{R_{L} + R_{2}}) I = (\frac{1.5}{1.5 + 6}) \times 7.5 = 1.5 mA$

$∴$ $I_{R_{L}} = I_{R_{1}} - I_{R_{2}} = 7.5 - 1.5 = 6 mA$

$(2) Potential difference across load$

$V_{R_{L}} = (I_{R_{L}}) R_{L} = 6 \times 1.5 = 9 V$

$(3) Ratio of powers dissipated in R_{1} and R_{2}$

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

$(4) When R_{1} and R_{2} are interchanged, then$

$R'_{p} = \frac{R_{2} R_{L}}{R_{2} + R_{L}} = \frac{2 \times 1.5}{2 + 1.5} = \frac{6}{7} k Ω$

$R'_{total} = 6 + R'_{p} = 6 + \frac{6}{7} k Ω$

$Now p.d. across R_{L}$

$V'_{R_{L}} = 24 (\frac{6 / 7}{6 + 6 / 7}) = 3 V$

$i.e., now potential becomes \frac{3}{9} = \frac{1}{3} rd .$

$Therefore power dissipated$ $P = \frac{V^{2}}{R} or P \propto V^{2}$ $will decrease by a factor of 9 .$

Solution

Q.
For the circuit shown in the figure [2009]

1 the current $I$ through the battery is $7.5 mA$

2 the potential difference across $R_{L}$ is 18 V

3 ratio of powers dissipated in $R_{1}$ and $R_{2}$ is 3

4 if $R_{1}$ and $R_{2}$ are interchanged, magnitude of the power dissipated in $R_{L}$ will decrease by a factor of 9

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Solution

Q. For the circuit shown in the figure [2009]

1 the current I through the battery is 7.5 mA

2 the potential difference across RL is 18 V

3 ratio of powers dissipated in R1 and R2 is 3