Problem 31.61 Part A For the circuit shown in the figure(Figure 1) find the curr

Question

Problem 31.61

Part A

For the circuit shown in the figure(Figure 1) find the current through each resistor.

Express your answers using two significant figures. Enter your answers numerically separated by commas.

Part B

For the circuit shown in the figure find the potential difference across each resistor.

Express your answers using two significant figures. Enter your answers numerically separated by commas.

Figure 1 of 1

Problem 31.61

Part A

For the circuit shown in the figure(Figure 1) find the current through each resistor.

Express your answers using two significant figures. Enter your answers numerically separated by commas.

I3, I24, I5, I4, I12 = A

Part B

For the circuit shown in the figure find the potential difference across each resistor.

Express your answers using two significant figures. Enter your answers numerically separated by commas.

V3, V24, V5, V4, V12 = V

Figure 1 of 1

24 2 4 5 n WW 12 V 12 (2 3 V WW 3 n

Explanation / Answer

here 4 ohm and 12 ohm resistance are in parellel

so the equilent resistance

= (4x12) / (4+12)

= 3 ohm

now 3 ohm and 5 ohm are in series

so 3 + 5 = 8 ohm

now 8 ohm and 24 ohm are in parallel

so the equivalent resistance = (8x24) / (8+24)

= 6 ohm

now apply KVL in loop

12 - 6I - 3 - 3I = 0

9I = 9

I = 1 Amp

so I(3 ohm) = I(6 ohm) = 1 Amp

so voltage V(3 ohm) = 1 x 3 = 3 volt

6 ohm is the resultant of parallel combination of 8 ohm and 24 ohm

so voltage across 6 ohm V(6 ohm) = 1 x 6 = 6 volt

voltage is same in parallel combination so

V(8 ohm) = V(24 ohm) = 6 volt

so I(8 ohm) = 6 / 8 = 0.75 Amp

and I(24 ohm) = 6 / 24 = 0.25 Amp

8 ohm is the resultant of series combination of 3 ohm and 5 ohm

current is same in series combination so

I(3 ohm) = I(5 ohm) = 0.75 Amp

so the voltage V(3 ohm) = 0.75 x 3 = 2.25 volt

and V(5 ohm) = 0.75 x 5 = 3.75 volt

3 ohm resistance is the resultant of parallel combination of 4 ohm and 12 ohm

so voltage is same in parallel so

V(4 ohm) = V(12 ohm) = V(3 ohm) = 2.25 volt

so current I(4 ohm) = 2.25 / 4 = 0.5625 Amp

and I(12 ohm) = 2.25 / 12 = 0.1875 Amp

(a) I(3 ohm) = 1 Amp I(24 ohm) = 0.25 Amp I(5 ohm) = 0.75 Amp I(4 ohm) = 0.5625 Amp

I(12 ohm) = 0.1875 Amp

(b) V(3 ohm) = 3 volt V(24 ohm) = 6 volt V(5 ohm) = 3.75 volt V(4 ohm) = 2.25 volt

V(12 ohm) = 2.25 volt

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