The circuit in the figure shows three identical lightbulbs attached to an ideal

Question

The circuit in the figure shows three identical lightbulbs attached to an ideal battery with epsilon = 20V. If the resistance of each bulb is 15 ohm, find the current through bulb 1 and the current through bulb 2. A particle of charge 30 mu C and mass 1.4 times10^-16 kg moves at a speed of 2000m/s, and its velocity vector lies g in the x-y plane. There is a magnetic field of magnitude 0.4 T along the +z direction. The magnetic force on the particle causes the particle to move in a circle that lies in the x-y plane. Find the radius of the circle and period of motion. Is the particle moving clockwise or counter-clockwise?

Explanation / Answer

Emf of the battery E = 20 volt

Bulb 2 and 3 are in parallel.Resultant of these two is R = (15x15)/(15+15)

R = 7.5 ohm

7.5 ohm and bulb 1 are in series.

Equivalent resistance R ' = 7.5 ohm + 15 ohm = 22.5 ohm

Current in the circuit i = E / R ' = 20 / 22.5

= 0.8888 A

Current through bulb1 is = 0.8888A

Current through bulb 2 is = 0.8888A/2 = 0.4444A

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