The resistance between points a and b in figure P18.52 drops to one-half its ori

Question

The resistance between points a and b in figure P18.52 drops to one-half its original value when switch S is closed.

Suppose a 32 V battery (with no internal resistance) is inserted between points a and b. What is the current through the battery before the swtich is closed (IBEFORE) and after the switch is closed (IAFTER)?

Question options:

IBEFORE = 2 A and IAFTER = 1 A

IBEFORE = 1 A and IAFTER = 2 A

IBEFORE = 0.5 A and IAFTER = 1 A

IBEFORE = 1 A and IAFTER = 0.5 A

IBEFORE and IAFTER can not both be determined with the information stated.

IBEFORE = 2 A and IAFTER = 1 A

IBEFORE = 1 A and IAFTER = 2 A

IBEFORE = 0.5 A and IAFTER = 1 A

IBEFORE = 1 A and IAFTER = 0.5 A

IBEFORE and IAFTER can not both be determined with the information stated.

Explanation / Answer

When Switch is open,

Req = R + (100*100)/(100+100)
Req before = R + 50

When switch is closed,
Req = R + (90*10)/(90+10) +  (90*10)/(90+10)
Req after = R + 18

Now as the question says,
Req after = 1/2 * Req before
R + 18 = 1/2 * (R + 50)
R = 14

So Req before, = 14 + 50 = 64 ohm
Req after, = 14 + 18 = 32 ohm

Now,
V = 32.0 V
V = I*R

Ibefore = V/Rbefore
Ibefore = 32.0/64
Ibefore = 0.5 A

Iafter = V/Rafter
Iafter = 32.0/32
Iafter = 1.0 A

Correct Option - (c)

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