a). Using series/parallel resistance reductions, find the equivalent resistance

Question

a). Using series/parallel resistance reductions, find the equivalent resistance between terminals A and B in the circuit of the figure below.

b). Repeat, but with C and D shroted together.

a). Using series/parallel resistance reductions, find the equivalent resistance between terminals A and B in the circuit of the figure below. b). Repeat, but with C and D shroted together.

Explanation / Answer

a) 10+30(series) and 30+10(series) are in parallel with 60 and each other => eq resistance = 1/60 + 1/40 +1/40 = 15 ohms This is in series with 5 and 10 ohms => Net eq resistance = 5+10+15 = 30 ohms b) here c and d are shorted then 10 and 30(parallel) and 30 & 10(parallel) are in series => eq resistance = 10*30/(10+30) + 10*30/(10+30) = 15 ohms This 15 ohms is in parallel to 60 ohms => eq resistance = 1/15 + 1/60 = 12ohms this 12 ohms is in series with 5 and 10 ohms Net eq resistance = 12+5+10 = 27 ohms

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