Two light bulbs (R_A = R_B = 10 Ohm) and two resistors are connected to an ideal

Question

Two light bulbs (R_A = R_B = 10 Ohm) and two resistors are connected to an ideal battery (Delta V = 9 V), as shown in the figure. Assume all values are known to three significant figures. (a) Initially, both bulbs A and B are glowing. Calculate the current flowing out of the battery, and the potential difference Delta V_12 = V_1 - V_2 between points 1 and 2. (b) Which bulb is brighter, A or B? Justify your answer. (c) Bulb B is removed from its socket (so its resistance R_B rightarrow infinity). What is Delta V_12 = V_1 - V_2 now? Explain your reasoning.

Explanation / Answer

let's find the equivalent resistance of circuit = (20x 15)/ 35 =8.57 ohms apprx

a) I = 9/8.57 = 1.05 A, current through right hand branch = 1.05 (20/ 35) = 0.6 A

V12= 0.6 ( 10 ) = 6 V apprx

b)Bulb is brighter as more current flows through bulb B

c)if bulb B is removed, it will behave like open circuit with infinite resistance and voltage drop will develop across it

V12=9 V

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