Two bulbs are connected in parallel across a source of EMF = 8.0V with a negligible internal resistance. One bulb has a resistance of 3.0Ω , and the other is 2.5Ω . A resistor R is connected in the circuit in series with the two bulbs.
(a) What value of R should be chosen in order to supply each bulb with a voltage of 2.4 V?
(b) For what value of R would the potential difference across each of the bulbs be 2.4 V?

(a) 3.18Ω

(b) 3.18Ω

Let the two bulbs be A and B

Given;

= Resistance in bulb A = 3.0Ω

= Resistance in bulb B = 2.5Ω

Since the two bulbs are connected in parallel;

i. their effective resistance () is given by

= +  ---------------(i)

Substitute the values of and into equation (i)

=> = +

Solve for  

= 1.36Ω

ii. voltage (potential difference), V, across them is the same;

Therefore we can get the total current (I) that will flow through them if the voltage to be supplied is 2.4V.

Use the Ohm's law;

V = I x R    -----------------(ii)

Where;

V = voltage across them = 2.4V

I = total current flowing through them

R = their effective resistance = = 1.36Ω

Substitute these values into equation (ii) as follows;

2.4  = I x 1.36

I = 2.4 / 1.36

I = 1.76A

(a) Now get the value of R

Since the voltage across the two bulbs is 2.4V out of the 8.0V supplied by the source, then the remaining (8.0 - 2.4 = 5.6)V will pass across the resistor R.

Also, since the two bulbs make a series connection with the resistor R, the same total current (I = 1.76A) that flows through these bulbs will flow through the resistor R.

Therefore, to get the value of R, we use the relation

V = I x R   ------------------------------(iii)

Where;

V = voltage across the resistor = 5.6V

I = current through the resistor = 1.76A

Substitute these values into equation (iii)

=> 5.6 = 1.76 x R

=> R  = 5.6 / 1.76

=> R = 3.18Ω

Therefore, the value of R to be chosen in order to supply each bulb with a voltage of 2.4V is 3.18Ω

(b) The potential difference and voltage across refer to the same thing. Therefore, the value of R that would make the potential difference across each of the bulbs be 2.4V is the same as the one calculated in (a) which is 3.18Ω

increase masses, decrease separation