d
(d) \({R_1} = 80 \times 200 = 16000\,\Omega = 16\,k\Omega \)
Current flowing through \({V_1}\)\(=\) Current flowing through \({V_2}\) \(=\) \(\frac{{80}}{{16 \times {{10}^3}}} = 5 \times {10^{ - 3}}\,A\).
So, potential differences across \({V_2}\) is
\({V_2} = 5 \times {10^{ - 3}} \times 32 \times {10^3} = 160\,{\rm{volt}}\)
Hence, line voltage \(V = {V_1} + {V_2} = 80 + 160 = 240\,V\).
