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Q.
Determine the potential difference between the points $C$ and $D$ in the given figure.
Current Electricity
Solution:
Step 1 : Main current I = $\frac{12}{(1 + 6 +3)} $ = 1.2 A
Potential difference across AB= 12 - 1.2 $\times$ l = 10.8 V $(terminal \,voltage \, V_{tmnl})$
Potential difference across AD= 6 $\times$ 1.2 = 7.2 V
Step 2 : . Potential difference across AC = $\frac{q}{C}$ ...$(i)$
Capacitors are in series $\therefore $ $q$ is same or $q = V_{tmnl} \times C_{eq}$
Here, $\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} = \frac{2 + 1}{2}$ or $C_{eq} = \frac{2}{3} \mu$F
$\therefore $ $q = 10.8 \times \frac{2}{3} \times 10^{-6} = 7.2 \times 10^{-6}$ C
Putting in equation $(i)$, Potential difference across AC = $\frac{7.2 \times 10^{-6}}{2 \times 10^{-6}} $ = 3.6 V
Step 3 : Potential difference between C and D is 7.2 - 3.6 = 3.6 V
SHort CuT Terminal voltage of 10.8V is available across series combination of capacitors of l$\mu$F and 2$\mu$F in the ratio of 2 : 1 giving 3.6V across 2$\mu$F capacitor. Similarly terminal voltage of 10.8 V is available to series combination of 3$\Omega$ and 6$\Omega$ in the ratio of 1 : 2 giving 7.2 V across 6$\Omega$ resistor