# Consider the circuit shown in the figure below. (R = 27.0 Ω.)?

Figure found at this site:

http://www.webassign.net/pse/p28-09alt.gif

(a) Find the current in the R = 27.0 Ω resistor.

??A

(b) Find the potential difference between points a and b.

??V

### 2 Answers

- MadhukarLv 710 years agoFavorite Answer
R and 5 Ω are in series

=> their equivalent resistance = 27 + 5 = 32 Ω

This 32 Ω, 5 Ω and 10 Ω are in parallel.

If R' is their equivalent resistance,

1/R' = 1/10 + 1/5 + 1/32

=> R' = 320 / (32 + 64 + 10) = 3.02 Ω.

R' is in series with 10 Ω

=> circuit resistance = 10 + 3.02 = 13.02 Ω

=> Main line current = 25/13.02 A = 1.92 A

(b)

P.D. across a and b

= 25 - p.d. across 10 Ω

= 25 - 1.92 * 10

= 5.8 volt

(a)

Current in R

= p.d. across R + 5 Ω branch / (R + 5)

= 5.8 / (27+5)

= 0.18125 A.

- rudicilLv 43 years ago
R + 5 = 33 ohms this is in parallel with 10 and 5, for a entire of a million/R = a million/33 + a million/10 + a million/5 R = 3.03 ohms upload the suitable 10 to get 13.03 ohms entire entire curren tis 25/13.03 = a million.919 amps that motives a voltage acros the parallel trio, that's Vba, of E = IR = a million.919 x 3.03 = 5.80 one volts., that's Vba cutting-edge interior the decrease branch is 5.80 one / 33 = 0.176 amps examine the maths