Question

1. Three 1500 Ω resistors are wired in parallel with a 15 V battery. What is the current through each of the resistors?
2. Three 1500 Ω resistors are wired in parallel with a 15 V battery. What is the current through the battery?
3. Can any arbitrary combination of resistors be broken down into series and parallel combinations?
4. Three resistors (110 Ω, 2500 Ω, and 5800 Ω) are wired together in series with a 12 V battery. What is the current through each of the resistors?
5. Three resistors (110 Ω, 2500 Ω, and 5800 Ω) are wired together in parallel with a 12 V battery. What is the current through each of the resistors?

Answer 1

(a) Three resistors R = 1500 ohm are connected in parallel.

So, the equivalent resistance, Req = R/3 = 1500 / 3 = 500 ohm

Since the resistances are connected in parallel across the battery so, the potential difference across each the resistance will be equal to the voltage of the battery.

So, current through each resistance = V/R = 15 / 1500 = 0.01 A

(b) Current through battery, I = V / Req = 15 / 500 = 0.03 A

(c) Yes. We can arrange any arbitrary combination of resistors in series and parallel according to our requirement.

(d) The resistors are connected in series.

Hence, equivalent resistance, Req = 110 + 2500 + 5800 = 8410 ohm

Voltage of the battery = 12 V

In series connection, current through each resistance will be the same.

And, the magnitude of the current, I = V / Req = 12 / 8410 = 0.00143 A (Answer)

(e) The resistances are connected in parallel.

Find out the equivalent resistance of the circuit.

110 ohm and 2500 ohm are in parallel.

So, its equivalent = (110*2500) / (110+2500) = 105.4 ohm

this combination is in parallel with 5800 ohm resistance.

Therefore, total equivalent resistance of the circuit, Req = (105.4*5800) / (105.4+5800) = 103.5 ohm

Hence, total current through the battery, I = V / Req = 12 / 103.5 = 0.116 A

So, current through 5800 ohm, I(5800) = 0.116*105.4 / (105.4+5800) = 0.002 A

total current through the combination of 110 ohm and 2500 ohm = 0.116 - 0.002 = 0.114 A

So, current through 110 ohm, I(110) = 0.114*2500 / (2500+110) = 0.109 A

current through 2500 ohm, I(2500) = 0.114*110 / (2500+110) = 0.005 A

Therefore, our answers are -

current through 110 ohm, I(110) = 0.109 A

current through 2500 ohm, I(2500) = 0.0.005 A

current through 5800 ohm, I(5800) = 0.002 A