If E_1 = 4 V and E_2 = 12 V,find the curren | vedclass

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(C) The circuit consists of two cells $E_1 = 4 \ V$ and $E_2 = 12 \ V$ connected in opposition,and three resistors in series: an internal resistance o

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$0.8 \ A, 6.4 \ V$

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(C) The circuit consists of two cells $E_1 = 4 \ V$ and $E_2 = 12 \ V$ connected in opposition,and three resistors in series: an internal resistance of $1 \ \Omega$ (implied for the $12 \ V$ cell),an internal resistance of $1 \ \Omega$ (implied for the $4 \ V$ cell),and an external resistor of $8 \ \Omega$.Applying Kirchhoff's voltage law in the loop:$I = \frac{E_{net}}{R_{total}} = \frac{12 \ V - 4 \ V}{8 \ \Omega + 1 \ \Omega + 1 \ \Omega} = \frac{8 \ V}{10 \ \Omega} = 0.8 \ A$.The potential difference between points $P$ and $Q$ is the voltage drop across the $8 \ \Omega$ resistor:$V_{PQ} = I 	imes R = 0.8 \ A 	imes 8 \ \Omega = 6.4 \ V$.

If $E_1 = 4 \ V$ and $E_2 = 12 \ V$,find the current in the circuit and the potential difference between the points $P$ and $Q$ respectively.

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