Question

In the following circuit, what is the voltage across PQ?

• A. $\frac{5}{3}V$
• B. $\frac{10}{3}V$
• C. $\frac{11}{3}V$
• D. $\frac{8}{3}V$

Answer 1

Answer: (B)

$\frac{10}{3}\,V$

Answer 2

Let the voltage across PQ be $V$. For the upper branch (4 V source with 1 Ω resistor), the current is

$$I_1=\frac{4-V}{1}.$$

For the lower branch (2 V source with 2 Ω resistor), the current is

$$I_2=\frac{2-V}{2}.$$

Since these branches are connected in parallel, in the absence of any external connection (only internal balancing), KCL gives

$$I_1 + I_2 = 0.$$

Thus,

$$\frac{4-V}{1}+\frac{2-V}{2}=0.$$

Multiply by 2:

$$2(4-V)+(2-V)=0 \quad \Longrightarrow \quad 8-2V+2-V=0.$$

Combine like terms:

$$10-3V=0 \quad \Longrightarrow \quad V=\frac{10}{3}\,V.$$

Explanation (minimal): Write branch currents as $(4-V)/1$ and $(2-V)/2$. Set their sum to zero, solve to get $V=\frac{10}{3}\,V$.