In the circuit shown PD across $$R$$ is $$V$$. Find the PD across the resistance $$2R$$
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PLACEHOLDER ANSWER - the answer is an expression in $$V$$. $$2R$$ and $$4R$$ in series ($$6R$$) are in parallel with $$3R$$, giving $$2R$$, which is in series with $$R$$. The same current flows through $$R$$ and the parallel group, so the PD across the group is $$2V$$. The $$6R$$ branch then carries $$\frac{2V}{6R}$$, and the PD across $$2R$$ is $$\frac{2V}{6R} \times 2R = \frac{2V}{3}$$.