12 Questions of Electricity and Magnetism
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(See attached file for full problem description with diagrams)
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a) Find the equivalent resistance of the circuit.
b) Find the current through the battery.
c) Find the voltage across the 800 ohm resistor.
c) Find the current in the 560 ohm resistor.
d) Find the voltage across the 400 ohm resistor.
Four identical light bulbs are connected to a voltage source as shown below.
a) Rank the light bulbs by brightness, from brightest to dimmest.
If any have the same brightness, explicitly say so.
Suppose Bulb B is unscrewed from its socket.
b) Does the current through the battery increase, decrease, or remain the same? Justify your answer.
c) Does bulb A get brighter, dimmer or remain the same? Explain.
d) Does bulb C get brighter, dimmer or remain the same? Explain.
Two wires carry currents as shown below.
a) What is the direction of the magnetic field due to the 5A wire at point P? (Draw an arrow in the space below to indicate the direction or indicate that it is into or out of the page).
b) What is the direction of the magnetic field due to the 15 A wire at point P? (Draw an arrow in the space below to indicate the direction or indicate that it is into or out of the page).
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Solution Summary
Electricity and magnetism questions are analyzed. The expert finds the equivalent resistance of a circuit and the current through the battery.
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Please see the attached file.
a) Find the equivalent resistance of the circuit.
600 and 400 resistances are in parallel and their equivalent is
given by
1. R1 = 600*400/(600 + 400) = 240
2. Now this resistance R1 in series with 560 resistor hence their equivalent is given by
R2 = 240 + 560 = 800
3. This R2 = 800 is in parallel with given 800 and hence their equivalent resistance or the resistance of the whole circuit is given as
R = 800*800/(800 + 800) = 400
b) Find the current through the battery.
As the equivalent resistance of the circuit is 400 the current through the Battery (considering internal resistance of the battery to be negligible) is given by
I = V/R = 12/400 = 0.03 A = 30 mA.
c) Find the voltage across the 800 ohm resistor.
As the ends of 800 resistor are directly connected to the battery terminals the voltage across it is same as the voltage across the terminals of the battery and is 12 Volts.
c) Find the current in the 560 ohm resistor.
The ends of the combination of 600, 400 and 560 are connected to the battery terminals, the voltage across it is 12 volts and the current through the combination is the ratio of voltage and the equivalent resistance of the combination R2 = 800 and hence current through the 560 resistance (same as the combination) is
I = 12/800 = 0.015 A = 15 mA.
d) Find the voltage across the 400 ohm resistor.
Voltage drop across 560 resistor will be (I*R) = 0.015*560 = 8.4 V
Hence the voltage drop across the parallel combination of 600 and 400 is the rest of the 12V which is 12 - 8.4 = 3.6 V
So the voltage across both and hence across 400 resistor = 3.6 ...
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