Problem 12-58 Resistor network - Part 6 - A
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.
\(I = I_{150} = V/R = 35/614 = 0.057 \;A \quad \text{(Answer to part (a))}\)
No. Only a portion of the \(35 \;V\) appears across the \(250 \Omega\) resistor, otherwise there would be \(0\) volts across the others. If that was so then from \(I = V/R\) there would be no current through them, but we just calculated that there was.