# GCSE Science/Parallel and series circuits answers

## Answers to practice questions on series circuits

edit*Q1)Let R*_{1}= 2Ω, R_{2}= 3Ω, and V=5V what is the voltage across R_{1}and R_{2}?- The total resistance =5Ω
- So the current is = 1A
- Apply Ohm's law to each resistor in turn gives:

- V
_{1}= 2V - V
_{2}= 3V

- V

*Q2)Let R*_{1}= 3Ω, R_{2}= 3Ω, and V=6V what is the voltage across R_{1}and R_{2}?- The total resistance is 6V
- The current is therefore 1A

- V
_{1}= 3V - V
_{2}= 3V

- V

*Q3)Let R*_{1}= 2Ω, V=5V and I= 1A what is the value of resistor R_{2}?- The total resistance = V/I

- =5/1
- = 5Ω

So R_{2} = 3Ω

## Answers to Questions on Parallel Circuits

edit*Q4)You disconnect one of the bulbs, does the other stay lit?*

Yes it does.

*Q5)You reconnect the disconnected bulb, does the other dim or brighten?*

Ideally it should not make any difference. The brightness of the lit bulb should remain the same. *(In practice the bulb will probably dim a tiny bit because of the internal resistance of the battery, but that's more advanced than this course)*

*Q6)Would it be reasonable to say "each branch of the circuit behaves as if the other branch weren't there" ?*

It certainly would.

*Q7) Calculate the resistance of the above pair of resistors*if*they had been in*series*rather than parallel.*

R_{1} + R_{2} = 1 + 1 = 2 Ω

*Q8) A parallel circuit has two resistors, one is 2 Ω the other is 3Ω what is the total resistance of the circuit?*

Let the Voltage be **V**.

From Ohm's Law

- current I
_{1}= V/2 A - current I
_{2}= V/3 A

So the total Current I = V/2+V/3

- = V(1/2 +1/3)
- = V (5/6)

Now we Apply Ohm's Law to the circuit again only this time using the total current so that we know what the total resistance is-

R = V/I

- = V/{V(5/6)}
- = 1/(5/6)
*We cancel the V's* - = 6/5
*To divide by a fraction, turn it upside down and multiply*

*Q9) Three resistors are in parallel (three separate branches) their resistances are 2Ω,2Ω and 3 Ω what is the resistance of the circuit?*

Rather than going through the whole procedure for Q8 again we note that -

1/R = 1/R_{1} + 1/R_{2} + 1/R_{3}

- = 1/2 + 1/2 + 1/3
- = 4/3

So R = 3/4 Ω