Series & Parallel circuit + bulb brightness

Nov 2015
38
0
Hello I'm working through a book (with answers) but am struggling with voltage, current, resistance and circuits. Please check my understanding below and let me know if I've finally understood. Thank you.
In particular I'm confused in Q2

Q1. A student connects light bulbs, A and B, and parallel across a battery. Bulb A is brighter than bulb B. Which bulb has the higher resistance? Explain.
ANSWER IN BOOK IS BULB B

Because it is a parallel circuit the same voltage MUST pass across each bulb i.e. there is a equal PUSH of current through each bulb. [I've learnt to think of voltage as a PUSH]

Bulb B is dimmer because it has a higher resistance. For the same voltage (PUSH) it resists the passage of current (electrons) so remains dim.
Bulb A is brighter because it has a lower resistance. For the same voltage (PUSH) it does not resist the passage of current (because it has low resistance) so becomes bright.

The relevant equation is P =IV
Say for Bulb A the current passing through is 6 A with a voltage (PUSH) of 10 V so the power the bulb receives to light the bulb is P=IV P = 6 x 10 P = 60 watts [THAT'S A BRIGHT BULB]

Say for Bulb B the current passing through is only 1 A (because of its higher resistance) with a voltage (PUSH) of 10 V so the power the bulb receives to light the bulb is P=IV P = 1 x 10 P = 10 watts [THAT'S A DIM BULB]

Q2. If the student connects the same bulbs in series with the same battery, which bulb will be brighter? Explain.
ANSWER IN BOOK IS BULB B will be brighter.

Because it is a series circuit the same current MUST pass through each bulb. The voltage across the components is shared in the ratio of the resistance of the components. The component with the higher resistance must have more voltage (PUSH) passing across it to deliver the same current.

Bulb B has the higher resistance so requires a high voltage across it (giving a big PUSH to the electrons) to allow current (electrons) to pass through it.

Bulb A has the low resistance so requires a low voltage across it to allow the same current (electrons) to pass through it.
In Bulb B with large voltage and the same current there will be large energy so the bulb will be brighter.
The relevant equation is P =IV

Say for Bulb B the current passing through is 10 A with a voltage (PUSH) of 100 V so the power the bulb receives to light the bulb is P=IV P = 10 x 100 P = 1000 watts [THAT'S A BRIGHT BULB]

For Bulb A the current passing through is also 10 A (because it is a series circuit) but a voltage (PUSH) of only 10 V is required (because it is a bulb of low resistance) so the power the bulb receives to light the bulb is P=IV P = 10 x 10 P = 100 watts [THAT'S A DIM BULB]
 
Apr 2015
1,035
223
Somerset, England
Yes your thinking is correct.

You can also get to answer 2 knowing that the power is I^2R so given the same current and therefore the same current, the higher resistance will absorb the greater power in the series circuit.

Can you think of a similar relationship that would help with answer 1 where the voltage is common?
 
Last edited:
Apr 2017
525
130
Just that you went a long way round to get the answer to question 2

The current is the same in each bulb ....power in each bulb is I squared R (IxIxR) , so the bulb with the highest resistance is brightest (consumes more power)...