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(B) The terminal potential difference $V$ is given by $V = E - Ir$,where $E$ is the emf and $r$ is the internal resistance. Also,$V = IR$,so $I = V/R$

Vedclass · Accepted Answer

$2 	ext{ V}, 1 \Omega$

Vedclass · Answer

(B) The terminal potential difference $V$ is given by $V = E - Ir$,where $E$ is the emf and $r$ is the internal resistance. Also,$V = IR$,so $I = V/R$.Case $1$: $V_1 = 1.6 	ext{ V}$,$R_1 = 4 \Omega$. The current $I_1 = 1.6 / 4 = 0.4 	ext{ A}$.Using $E = V_1 + I_1 r$,we get $E = 1.6 + 0.4r$ (Equation $i$).Case $2$: $A$ second $4 \Omega$ resistor is connected in parallel,so the equivalent resistance $R_2 = (4 	imes 4) / (4 + 4) = 2 \Omega$. The new potential difference $V_2 = 1.33 	ext{ V}$.The new current $I_2 = V_2 / R_2 = 1.33 / 2 = 0.665 	ext{ A}$.Using $E = V_2 + I_2 r$,we get $E = 1.33 + 0.665r$ (Equation $ii$).Equating $(i)$ and $(ii)$: $1.6 + 0.4r = 1.33 + 0.665r$.$1.6 - 1.33 = 0.665r - 0.4r \Rightarrow 0.27 = 0.265r \Rightarrow r \approx 1 \Omega$.Substituting $r = 1 \Omega$ into $(i)$: $E = 1.6 + 0.4(1) = 2 	ext{ V}$.Thus,the emf is $2 	ext{ V}$ and the internal resistance is $1 \Omega$.

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