Problem 12-58 Resistor network - Part 8 - C
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.
\(I = I_{150} = V/R = 35/614 = 0.057 \;A \quad \text{(Answer to part (a))}\)
\(\Delta V_{250} = (0.057 \;A)(250 \Omega) = 14.3 \;V \quad \text{(Answer to part (b))}\)
\(\Delta VR' = (0.057\; A)(214 \Omega) = 12.2\; V\)
No. A current goes through the \(350 \Omega\) resistor so there must be a potential across it.