This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Step 1: Calculate the resistance of each bulb. The power P of a bulb is related to its rated voltage V and resistance R by the formula P = (V^2)/(R). Therefore, R = (V^2)/(P). For the 100 W bulb (Bulb A): R_A = (200\, V)^2100\, W = 40000\, V^2100\, W = 400\, For the 60 W bulb (Bulb B): R_B = (200\, V)^260\, W = 40000\, V^260\, W = (2000)/(3)\, ≈ 666.67\, Step 2: Calculate the total resistance of the series circuit. When bulbs are connected in series, the total resistance is the sum of individual resistances. R_total = R_A + R_B = 400\, + (2000)/(3)\, = (1200)/(3)\, + (2000)/(3)\, = (3200)/(3)\, Step 3: Calculate the total current flowing through the circuit. The total current I in the series circuit is given by Ohm's law, I = V_lineR_total, where V_line is the supply voltage (200 V). I = 200\, V(3200)/(3)\, = (200 × 3)/(3200)\, A = (600)/(3200)\, A = (3)/(16)\, A Step 4: Calculate the potential drop across each bulb. The potential drop across each bulb is given by V = I · R. Potential drop across the 100 W bulb (V_A): V_A = I · R_A = (3)/(16)\, A × 400\, = (1200)/(16)\, V = 75\, V Potential drop across the 60 W bulb (V_B): V_B = I · R_B = (3)/(16)\, A × (2000)/(3)\, = (2000)/(16)\, V = 125\, V Step 5: Compare the potential drops. We found that V_A = 75\, V and V_B = 125\, V. This means the potential drop across the 100 W bulb (75\, V) is smaller than the potential drop across the 60 W bulb (125\, V). Comparing this with the given options: (A) Each bulb is 200 V - Incorrect. (B) 100 W bulb is greater than that across 60 W bulb - Incorrect. (C) 100 W bulb is smaller than that across 60 W bulb - Correct. (D) Each bulb is 100 V - Incorrect. The final answer is C Send me the next one 📸