Consider the circuit shown in(Figure 1) . The batteries have emfs of E1=9.0V and

Question

Consider the circuit shown in(Figure 1) . The batteries have emfs of E1=9.0V and E2=12.0V and the resistors have values of R1=28?, R2=64?, and R3=38?.

A. Determine the magnitudes of the currents in each resistor shown in the figure. Ignore internal resistance of the batteries.

B. Determine the directions of the currents in each resistor. Ignore internal resistance of the batteries.

A. I1 left, I2 right, I3 up

B. I1 left, I2 right, I3 down

C. I1 right, I2 left, I3 down

D. I1 right, I2 left, I3 up

Consider the circuit shown in(Figure 1) . The batteries have emfs of E1=9.0V and E2=12.0V and the resistors have values of R1=28?, R2=64?, and R3=38?. A. Determine the magnitudes of the currents in each resistor shown in the figure. Ignore internal resistance of the batteries. B. Determine the directions of the currents in each resistor. Ignore internal resistance of the batteries. A. I1 left, I2 right, I3 up B. I1 left, I2 right, I3 down C. I1 right, I2 left, I3 down D. I1 right, I2 left, I3 up

Explanation / Answer

This a one node circuit. The node is just to the left of the 68?. Call it V. Ground the right side of R2. Now write equations for the currents leaving V and by KCL that sum = 0
[V-(-12)]/38 + V/64 + [V-(-9)]/28 = 0
V(1/38+1/64+1/28) = -12/38 - 9/28
V = -8.2
Let i1 be the current in R1, i2 in R2, i3 in R3
i1 = (V+9)/28 = 0.02857A UP
i2 = V/64 = -0.128A RIGHT or +0.128A LEFT
I3 = (V+12)/38 = 0.1A DOWN

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