Asked by brock
a parallel combination of two resistors of 2ohm each has third resistor of 4ohm in series with the combination.if a battery of 10V with negligible internal resistance be applies across them,find the p.d. across the resistor.
Answers
Answered by
drwls
Two 2 ohm resistors in parallel act like a single 1 ohm resistor in series. Total circuit resistance, seen by the battery, is
4 + 1 = 5 ohms.
80% of the p.d. of the circuit is across the 4 ohm resistor.
80% of 10V = ?
4 + 1 = 5 ohms.
80% of the p.d. of the circuit is across the 4 ohm resistor.
80% of 10V = ?
Answered by
henry2,
Given: E = 10V., R1 = 2 ohms, R2 = 2 ohms, R3 = 4 ohms.
Rt = (2*2)/(2+2) + 4 = 1 + 4 = 5 ohms.
I = E/Rt = 10/5 = 2A.
V3 = I*R3 = 2 * 4 = 8V.
V1 = V2 = E-V3 = 10-8 = 2V.
Rt = (2*2)/(2+2) + 4 = 1 + 4 = 5 ohms.
I = E/Rt = 10/5 = 2A.
V3 = I*R3 = 2 * 4 = 8V.
V1 = V2 = E-V3 = 10-8 = 2V.
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