Voltmeters $V_1$ and $V_2$ are connected in series across a $D.C.$ line. $V_1$ reads $80\, volts$ and has a per volt resistance of $200 \,\Omega$. $V_2$ has a total resistance of $32 \,kilo-ohms$. The line voltage is .............
Diffcult
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(d) ${R_1} = 80 \times 200 = 16000\,\Omega = 16\,k\Omega $
Current flowing through ${V_1}$$=$ Current flowing through ${V_2}$ $=$ $\frac{{80}}{{16 \times {{10}^3}}} = 5 \times {10^{ - 3}}\,A$.
So, potential differences across ${V_2}$ is
${V_2} = 5 \times {10^{ - 3}} \times 32 \times {10^3} = 160\,{\rm{volt}}$
Hence, line voltage $V = {V_1} + {V_2} = 80 + 160 = 240\,V$.
Current flowing through ${V_1}$$=$ 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 $V = {V_1} + {V_2} = 80 + 160 = 240\,V$.
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