The V = 10.10 V battery in the figure below is removed from the circuit and rein

Question

The V = 10.10 V battery in the figure below is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. (R = 9.30 .)

The V-10.10 V battery in the figure below is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. (R = 9.30 .) 200 V 1,00) 5.00V (a) Find the current in each branch. upper branch middle branch ower branch (b) Find the potential difference Vab of point a relative to point b.

Explanation / Answer

V = 10.10 V ; R = 9.3 Ohm

Applying Khirchoff loop rule in both lower and upper loop we get

(4 + 1) I2 + 9.3(I1 + I2) + 5 = 0

5I2 + 9.3I1 + 9.3I2 + 5 = 0

9.3 I1 + 14.3 I2 = 0 (1)

10.10 - (3 + 2)I1 - 9.3(I1 + I2) = 0

10.10 - 5 I1 - 9.3 I1 - 9.3 I2 = 0

-14.3 I1 - 9.3 I2 + 10.10 = 0

I1 = (10.10 - 9.3 I2)/14.3

putting this in (1)

9.3 (10.10 - 9.3 I2)14.3 + 14.3 I2 + 5 = 0

6.57 - 6.05 I2 + 14.3 I2 + 5 = 0

I2 = 1.4 A

I1 = 0.204 A

I3 = 1.604 A

Hence, I(upper) = 1.604 ; I(middle) = 1.4 A ; I(lower) = 0.204 A

PD across ab will be:

Vab = 3(1.604) + 4(1.4) = 10.412 Volts

Hence, Vab = 10.412 V

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