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The current ia in the circuit shown in Fig. is 2 mA.

Find (a) i0; (b) ig; and (c) the power delivered by the independent current source.

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a)

Voltage across resistor 1k: $V_{1k} = 2*1-^{-3}mA*1*10^3 \Omega = 2V$

Voltage across resistor 4k: $V_{4k} = 2*1-^{-3}mA*4*10^3 \Omega = 8V$

Voltage across resistor 3k: $V_{3k} = 2*1-^{-3}mA*3*10^3 \Omega = 6V$

Voltage across two parallel branches is the same, thus:

Voltage across resistor 2k: $V_{2k} = V_{1k} + V_{4k} + V_{3k} = 16V$

$I = V/R$, -> $i_0 = 16V/(3*10^3\Omega) = 8mA$

b)

$i_g = i_a + i_0 = 10mA$

c)

$P = I*V$, $P = 10mA * 16V = 0.16W$

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Edited Sat, 15 Apr '17
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