In the circuit shown,E, F, G,and H are cells | vedclass

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(D) Applying Kirchhoff's voltage law to the loops:For loop $AFDBEA$ (assuming current $i_1$ clockwise): $E_F - E_E = i_1(r_F + r_E + R) - R i_2$$1 - 2

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(D) Applying Kirchhoff's voltage law to the loops:For loop $AFDBEA$ (assuming current $i_1$ clockwise): $E_F - E_E = i_1(r_F + r_E + R) - R i_2$$1 - 2 = i_1(1 + 2 + 2) - 2 i_2 \Rightarrow 5 i_1 - 2 i_2 = -1 \dots(1)$For loop $CGDBHC$ (assuming current $i_2$ clockwise): $E_H - E_G = i_2(r_H + r_G + R) - R i_1$$1 - 3 = i_2(1 + 3 + 2) - 2 i_1 \Rightarrow -2 i_1 + 6 i_2 = -2 \dots(2)$Multiplying $(1)$ by $3$: $15 i_1 - 6 i_2 = -3 \dots(3)$Adding $(2)$ and $(3)$: $13 i_1 = -5 \Rightarrow i_1 = -5/13\, A$Substituting $i_1$ in $(1)$: $5(-5/13) - 2 i_2 = -1 \Rightarrow -25/13 + 1 = 2 i_2 \Rightarrow i_2 = -6/13\, A$Potential difference $V_D - V_B = R(i_2 - i_1) = 2(-6/13 - (-5/13)) = 2(-1/13) = -2/13\, V$.Potential difference across $G$: $V_G = E_G + i_2 r_G = 3 + (-6/13)(3) = 3 - 18/13 = 21/13\, V$.Potential difference across $H$: $V_H = E_H - i_2 r_H = 1 - (-6/13)(1) = 1 + 6/13 = 19/13\, V$.

In the circuit shown,$E, F, G$,and $H$ are cells of e.m.f. $2\,V, 1\,V, 3\,V$,and $1\,V$ respectively,and their internal resistances are $2\,\Omega, 1\,\Omega, 3\,\Omega$,and $1\,\Omega$ respectively.

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