(a) 3.18Ω
(b) 3.18Ω
Let the two bulbs be A and B
Given;
[tex]R_{A}[/tex] = Resistance in bulb A = 3.0Ω
[tex]R_{B}[/tex] = Resistance in bulb B = 2.5Ω
Since the two bulbs are connected in parallel;
i. their effective resistance ([tex]R_{X}[/tex]) is given by
[tex]\frac{1}{R_{X}}[/tex] = [tex]\frac{1}{R_{A} }[/tex] + [tex]\frac{1}{R_{B} }[/tex] ---------------(i)
Substitute the values of [tex]R_{A}[/tex] and [tex]R_{B}[/tex] into equation (i)
=> [tex]\frac{1}{R_{X}}[/tex] = [tex]\frac{1}{3.0}[/tex] + [tex]\frac{1}{2.5}[/tex]
Solve for [tex]R_{X}[/tex]
[tex]R_{X}[/tex] = 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 = [tex]R_{X}[/tex] = 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
