Question

The resistor \( R_1 = 30 \, \Omega \) and \( R_2 = 10 \, \Omega \) are connected in parallel to a 20 V battery. Find the heat developed in the resistor \( R_1 \) in one minute

• A. 600 J
• B. 800 J
• C. 6000 J
• D. 8000 J
• E. 7000 J

Hint:

For components connected in parallel, the voltage across them is the same. This makes the power formula \( P = V^2/R \) and heat formula \( H = (V^2/R)t \) particularly useful.

Answer 1

The Correct Option is B

Step 1: Identify the given information and the formula for heat developed.
Given: - Resistance of the first resistor, \( R_1 = 30 \, \Omega \). - Voltage of the battery, \( V = 20 \, \text{V} \). - Time, \( t = 1 \, \text{minute} \). The formula for heat (Joule's heating) developed in a resistor is given by: \[ H = I^2 R t = \frac{V^2}{R} t = V I t \] Since the voltage across the resistor is known, the formula \( H = \frac{V^2}{R} t \) is the most direct to use.

Step 2: Determine the voltage across resistor \( R_1 \).
The resistors \( R_1 \) and \( R_2 \) are connected in parallel to the battery. In a parallel circuit, the voltage across each component is the same as the voltage of the source. Therefore, the voltage across \( R_1 \) is: \[ V_1 = V = 20 \, \text{V} \]

Step 3: Convert time to SI units.
The time is given in minutes, which needs to be converted to seconds for consistency in SI units. \[ t = 1 \, \text{minute} = 60 \, \text{seconds} \]

Step 4: Calculate the heat developed in \( R_1 \).
Substitute the values of \( V_1 \), \( R_1 \), and \( t \) into the heat formula: \[ H_1 = \frac{V_1^2}{R_1} t \] \[ H_1 = \frac{(20 \, \text{V})^2}{30 \, \Omega} \times 60 \, \text{s} \] \[ H_1 = \frac{400}{30} \times 60 \] \[ H_1 = 400 \times \frac{60}{30} = 400 \times 2 \] \[ H_1 = 800 \, \text{J} \] The heat developed in resistor \( R_1 \) in one minute is 800 Joules.

Final Answer: (B) 800 J