Two resistors of 200 kOmega and 1 MOmega form | vedclass

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(C) Let the resistors be $R_1 = 200\ k\Omega = 0.2\ M\Omega$ and $R_2 = 1\ M\Omega$.The total resistance is $R_{eq} = R_1 + R_2 = 0.2\ M\Omega + 1\ M\

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(C) Let the resistors be $R_1 = 200\ k\Omega = 0.2\ M\Omega$ and $R_2 = 1\ M\Omega$.The total resistance is $R_{eq} = R_1 + R_2 = 0.2\ M\Omega + 1\ M\Omega = 1.2\ M\Omega$.The potential difference across the series combination is $V_{total} = V_1 - V_2 = 3\ V - (-15\ V) = 18\ V$.Using the potential divider formula,the voltage at the junction $V_J$ relative to the negative terminal $(-15\ V)$ is:$V_J - (-15\ V) = V_{total} 	imes \frac{R_2}{R_1 + R_2}$$V_J + 15\ V = 18\ V 	imes \frac{1\ M\Omega}{1.2\ M\Omega}$$V_J + 15\ V = 18 	imes \frac{1}{1.2} = 15\ V$$V_J = 15\ V - 15\ V = 0\ V$.

Two resistors of $200\ k\Omega$ and $1\ M\Omega$ form a potential divider circuit. If the potentials at the two ends are $+3\ V$ and $-15\ V$,the voltage at the junction will be ............... $V$.

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