In the circuit shown below the 4 bulbs are identical and have a resistance of 10

Question

In the circuit shown below the 4 bulbs are identical and have a resistance of 105 each. The battery supplies a voltage of 45 V. S1, S2, and S3 represent switches.

(a) S1 and S2 are closed while S3 is open. What is the current through each bulb?

(b) S3 is closed while S1 and S2 are open. What is the current through each bulb?

(c) S2 and S3 are closed while S1 is open. What is the current through each bulb?

(d) S1 and S3 are closed while S2 is open. What is the current through each bulb?

(e) All three switches are closed. What is the current through each bulb?

Explanation / Answer

(a) In this case, there is a short circuit from Bulb 1 to Bulb 2, so R-equiv is 210 ohms,
and the current through each of Bulbs 1 and 2 is 45 / 210 = 0.214 amps. No current through 3 and 4.

(b) In this case, Bulbs 1,3,4 are in series, R-equiv is 315 ohms; current through 1,3,4
is 45 / 315 = 0.142 amps. No current through Bulb 2.

(c) In this case, Bulbs 1 and 3 are in series with the parallel-resistance combination
of Bulbs 2 and 4.

bulb 2 and 4 will give R24 = 105*105 / (105+105) = 52.5 ohms

R-equiv for the whole circuit is 52.5 + 105+105 = 262.5, and the current
through Bulbs 1 and 3 is 45 / 262.5 = 0.171 amps;

voltage across R24 is V24 = 0.171*52.5 = 9 v

current is split through Bulbs 2 and 4,

current in bulb 2 and 4 is i2 = i4 = 9/105 = 0.085 amps each.

(d) In this case, Bulb 1 is in series with a parallel combination whose two branches are
(i) Bulb 2 and (ii) Bulbs 3 and 4 in series. R-equiv for the whole circuit is
105 + 1/(1/105+1/210) = 175 ohms.
Current through Bulb 1 is 45 V / 175 ohms = 0.257 amps.
The parallel branches split this current, with twice as much of it going through Bulb 2
as through the 3-4 series combo. So the current in Bulb 2 is 0.257*(2/3) = 0.171 amps, and the
current through Bulbs 3 and 4 is 0.0855 amps.

(e)

no current will go through Bulb 3,
because the only place you can go from there is to Bulb 2 or Bulb 4, both of
which can be reached from Bulb 1 with (theoretically) no resistance at all.

So I think this one is going to behave as if the parallel combination of Bulb 2 and 4
is in series with Bulb 1. Hence, R-equiv is 1.5 x 105 ohms = 157.5 ohms.
Current through Bulb 1 is 0.285 ohms,

currents through Bulbs 2 and 4 are 0.143 each

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