In the circuit shown,the currents i_{1} and i | vedclass

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(B) First,calculate the equivalent resistance of the parallel combination of $12 \Omega$ and $4 \Omega$ resistors:$R_p = \frac{12 	imes 4}{12 + 4} =

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$i_{1}=0.5 	ext{ A}, i_{2}=1.5 	ext{ A}$

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(B) First,calculate the equivalent resistance of the parallel combination of $12 \Omega$ and $4 \Omega$ resistors:$R_p = \frac{12 	imes 4}{12 + 4} = \frac{48}{16} = 3 \Omega$Now,calculate the total resistance of the circuit including the series resistor and internal resistance:$R_{total} = R_p + 2 \Omega + 1 \Omega = 3 \Omega + 2 \Omega + 1 \Omega = 6 \Omega$The total current $I$ flowing from the battery is:$I = \frac{V}{R_{total}} = \frac{12 	ext{ V}}{6 \Omega} = 2 	ext{ A}$Using the current divider rule for parallel resistors:$i_1 = I 	imes \frac{R_2}{R_1 + R_2} = 2 	imes \frac{4}{12 + 4} = 2 	imes \frac{4}{16} = 0.5 	ext{ A}$$i_2 = I 	imes \frac{R_1}{R_1 + R_2} = 2 	imes \frac{12}{12 + 4} = 2 	imes \frac{12}{16} = 1.5 	ext{ A}$Thus,$i_1 = 0.5 	ext{ A}$ and $i_2 = 1.5 	ext{ A}$.

In the circuit shown,the currents $i_{1}$ and $i_{2}$ are

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