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Experience:  Master of Computer Applications (MCA). BSc in Mathematics, Physics, and Computer Science.
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# physics -- three resistors of 8, 12 and 24 Watts are in parallel

### Resolved Question:

physics -- three resistors of 8, 12 and 24 Watts are in parallel and a current of 20 amps is drawn by the combination. Determine (a) the voltage across the combo (b) the current through each resistor. thx
Submitted: 5 years ago.
Category: Homework
Expert:  Chris Parker replied 5 years ago.
Hi!

Let the three resistors be R1, R2, and R3.

R1 =8Ω, R2 =12Ω, and R3=24Ω.

For parallel circuits, 1 / R = 1 / R1 + 1 / R2 + 1 / R3.

Let us plug the values. 1 / R = 1/8 + 1/12 + 1/24 = 6/24 =1/4.

R = 4Ω.

a) Voltage across the combo can be determined by the formula V = I R.

By plugging the values, we get V = 20 x 4 = 80V.

b) Current through each resistor can be measured as I1 = V/R1, I2 = V/R2, and I3 = V/R3.

By plugging the values, we get I1 = 80/8 =10A, I2 =80/12 =6.67A, and I3 =80/24 =3.33A.

Hope that helped. Please review and click Accept.

Regards,
Chris
Customer: replied 5 years ago.

But the three resistors are in WATTS not Ohms
Expert:  Chris Parker replied 5 years ago.
Ok. For parallel circuits, total power P = P1 + P2 + P3.

P1=8, P2 =12, P3 =24. P = 8 + 12 + 24 =44 Watts.

a) Voltage across the combo V= P/I. By plugging I=20A and P=44 Watts, we get V= 44/20 = 2.2V.
b) Current through each resistor can be measured as I1 = P1/V, I2 =P2/V, and I3 = P3/V.
By plugging the values, we get I1 = 8/2.2 =3.63A, I2 =12/2.2 =5.45A, and I3 =24/2.2 =10.9A.

Regards,
Chris
Customer: replied 5 years ago.

Chris, quick follow up question -- previously you mentioned "For parallel circuits, total power P = P1 + P2 + P3" -- Is this also true for Power in series or is the equation different? Thx
Expert:  Chris Parker replied 5 years ago.