Problem 12-58 Resistor network - Part 6 - B

In the Figure, what is the
(a) current through the \(150 \Omega\) resistor?
(b) potential difference across the \(250 \Omega\) resistor?
(c) current through the \(350\Omega\) resistor?
(d) potential at \(\text{point A}\) (assuming zero potential at the negative terminal of the battery)

[Ans. (a) \(0.057 \;A\)   (b) \(14 \;V\)   (c) \(0.035\; A \)  (d) \(21 \;V\) ]

Accumulated Solution

\(1/R' = 1/350 + 1/550 , \quad R' = 214 \Omega \\ R = 150 + 214 + 250 = 614 \Omega\)

Current flow is counter clockwise.

\(I = I_{150} = V/R = 35/614 = 0.057 \;A \quad \text{(Answer to part (a))}\)

\(\Delta V_{250} = (0.057 \;A)(250 \Omega) = 14.3 \;V \quad \text{(Answer to part (b))}\)

Correct!

The potential difference across the parallel pair (\(350 \Omega\) and \(550 \Omega\)) is the potential difference across \(R'.\) It is:

(A)   \(35\; V\)

(B)   \((0.0057 \;A)(214 \Omega) = 12.2 \;V\)

(C)   \(0 \;V\)