Find the current in the 10 , Omega resistance | vedclass

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(C) The circuit consists of two cells of $EMF$ $\varepsilon_1 = 5 \, V$ and $\varepsilon_2 = 5 \, V$ with internal resistances $r_1 = 2 \, \Omega$ and

Vedclass · Accepted Answer

$0.45 \, A$ from $P_2$ to $P_1$

Vedclass · Answer

(C) The circuit consists of two cells of $EMF$ $\varepsilon_1 = 5 \, V$ and $\varepsilon_2 = 5 \, V$ with internal resistances $r_1 = 2 \, \Omega$ and $r_2 = 2 \, \Omega$ connected in parallel across a load resistance $R = 10 \, \Omega$.The equivalent $EMF$ $\varepsilon_{eq}$ of the parallel combination is given by:$\varepsilon_{eq} = \frac{\frac{\varepsilon_1}{r_1} + \frac{\varepsilon_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{5}{2} + \frac{5}{2}}{\frac{1}{2} + \frac{1}{2}} = \frac{2.5 + 2.5}{0.5 + 0.5} = \frac{5}{1} = 5 \, V$The equivalent internal resistance $r_{eq}$ is given by:$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{2} + \frac{1}{2} = 1 \, \Omega^{-1} \Rightarrow r_{eq} = 1 \, \Omega$The total current $i$ flowing through the load resistance $R$ is:$i = \frac{\varepsilon_{eq}}{R + r_{eq}} = \frac{5}{10 + 1} = \frac{5}{11} \approx 0.45 \, A$Since the positive terminals of the batteries are connected to the $P_2$ side, the current flows from $P_2$ to $P_1$ through the $10 \, \Omega$ resistor.

Find the current in the $10 \, \Omega$ resistance.

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