Question 562: Current Electricity

Question

Potential difference between the points P and Q is nearly

Accepted Answer

As, the resistors across each branch are in series.
$$\begin{aligned}
& \mathrm{R}_1=6+3=9 \Omega \
& \mathrm{R}_2=8+4=12 \Omega
\end{aligned}$$
According to $$\mathrm{KCL}$$, the current (i) will get divided into two parts $$I_1$$ and $$I_2$$
$$I_1=\frac{R_2}{\left(R_1+R_2ight) i}=\frac{12}{9+12} 	imes 5=2.86 \mathrm{~A}$$
Potential difference between $$P$$ and $$Q$$ is $$V=I_1 R=2.86 	imes 6=17.14 \mathrm{~V}$$

Answer

6 V

Answer

8 V

Answer

17 V

Answer

21 V

Current Electricity

Potential difference between the points P and Q is nearly

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