Two resistors A and B have resistance 5 ohm and 10 ohm, respectively. If they are connected in series to a voltage source of 5 V. The ratio of power developed in resistor A to that of power developed in resistor B will be:

Correct option is B

The power dissipated in a resistor (P) is given by:

P=I²R

In the case of resistors in series, the current I remains the same for both resistors.

Given Data:

• Resistance of A(R_A)=5 Ω
• Resistance of B(R_B)=10 Ω
• Voltage of the source (V) = 5 V

Step-by-Step Solution:

• Current in the Circuit: The total resistance in series is:

R_total=R_A+R_B=5+10=15 Ω

The current (I) in the circuit is:

​$$I= \frac{V}{R_{\text{total}}} = \frac{5}{15} = \frac{1}{3} \, \text{A}$$​​

• Power Dissipated in Resistors: The power dissipated in each resistor is:

P=I²R

• Power in A(P_A):

​$$P_A= I^2 R_A = \left(\frac{1}{3}\right)^2 \cdot 5 = \frac{1}{9} \cdot 5 = \frac{5}{9} W$$​​

• Power in B(P_B):

​$$PB = I^2 R_B = \left(\frac{1}{3}\right)^2 \cdot 10 = \frac{1}{9} \cdot 10 = \frac{10}{9} \, W$$​​

• Ratio of Power Dissipation: The ratio of P_A to P_B is:

​$$\frac{P_A}{P_B} = \frac{\frac{5}{9}}{\frac{10}{9}} = \frac{5}{10} = \frac{1}{2}$$​​

Final Answer:

The ratio of power dissipated in resistor A to that in resistor B is 1:2.