Physics - Current Electricity Question with Solution | TestHub

$R_{\text{total }}=2+\frac{6\times 1.5}{6+1.5}=3.2\mathrm{k}\Omega$ (A) $I=\frac{24\mathrm{V}}{3.2\mathrm{k}\Omega }=7.5\mathrm{mA}=I_{R_1}$ $I_{R_2}=\left(\frac{R_L}{R_L+R_2}\right)I$ $I=\frac{1.5}{7.5}\times 7.5=1.5\mathrm{mA}$ (B) $V_{R_L}=\left(I_{R_L}\right)\left(R_L\right)=9\mathrm{V}$ (C) $\frac{P_{R_1}}{P_{R_2}}=\frac{\left(I_{R_1}^2\right)R_1}{\left(I_{R_2}^2\right)R_2}=\frac{(7.5{)}^2(2)}{(1.5{)}^2(6)}=\frac{25}{3}$ (D) When $R_1$ and $R_2$ are interchanged, then $$\frac{R_2{R}_L}{R_2+R_L}=\frac{2\times 1.5}{3.5}=\frac{6}{7}\mathrm{k}\Omega$$ Now potential difference across $R_L$ will be $$V_L=24\left[\frac{6/7}{6+6/7}\right]$$ Earlier it was $9\mathrm{V}$ 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.