In the circuit above, the battery of emf epsilon is connected to two long, strai

Question

In the circuit above, the battery of emf epsilon is connected to two long, straight, parallel wires, which in turn are connected to four resistors with resistances given in the figure above. Assume that any other resistances in the circuit are negligible. Express all algebraic answers to the following parts in terms of the given quantities and fundamental constants. a. Derive an expression for the total resistance of the circuit. b. Derive an expression for the power dissipated in this circuit. c. How would connecting the R_2 resistors in series rather than parallel change the amount of current flowing through the circuit? Justify your answer. d. How does the amount of current flowing through R_2 compare to the amount of current flowing through R_1? Justify your answer. e. Given that R_2 = 2R_1, find the resistance of R_1 if the battery voltage is 10 volts and the total power dissipated by the circuit is 200 Watts.

Explanation / Answer

part a:

R2 and R2 are in parallel.

equivalent resistance=R2*R2/(R2+R2)=R2/2

it is in series with two R1.

so net resistance of the circuit=2*R1+0.5*R2

part b:

power dissipated=voltage^2/total resistance

=E^2/(2*R1+0.5*R2)

part c:

if R2 are connected in series, total resistance

=2*R2+2*R1

as total resistance increases, current will decrease.

part d:

using current division,

current through R2=current through R1*(R2/(R2+R2))

=0.5*current through R1

part e:

given R2=2*R1

E=10 volts

power =200 W

==>E^2/(2*R1+0.5*R2)=power

==>10^2/(2*R1+0.5*2*R1)=200

==>3*R1=10^2/200=0.5

==>R1=0.167 ohms

so resistance of R1 is 0.167 ohms

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