Question

Resistor A has twice the resistance of resistor B. The two are connected in series and a potential difference is maintained across the combination. The rate of thermal energy dissipation in A is: 1. one fourth that in B 2. the same as that in B 3. four times that in B 4. half that in B

Answer 1

Answer:

The thermal energy dissipated in A would be twice that in B

Explanation:

Resistor B (RB)= R

Resistor A (RA)= 2 R

When they are connected in series the equivalent Resistance in the circuit would be;

Equivalent resistance = RA +RB = R + 2 R = 3 R;

From ohms law I = V/R

I = V/3 R

Now the thermal energy is the power dissipated by the circuit and can be obtained thus;

P =$I^{2}R$

Then,

$P_{A} = (\frac{V}{3R}) ^{2} *2 R\\\\P_{A} = \frac{V^{2} }{9R^{2} } *2R\\\\P_{A} = \frac{2}{9}( \frac{V^{2} }{R}) \\\\P_{B} = (\frac{V}{3R}) ^{2} * R\\\\P_{B} = \frac{V^{2} }{9R^{2} } *R\\\\P_{B} = \frac{1}{9}( \frac{V^{2} }{R})$

Therefore Pa : Pb = 2: 1, this means that the thermal energy dissipated in A would be twice that in B