A battery with = 5.00 V and no internal resistance supplies current to the circu

Question

A battery with = 5.00 V and no internal resistance supplies current to the circuit shown in the figure below. When the double-throw switch S is open as shown in the figure, the current in the battery is 1.00 mA. When the switch is closed in position a, the current in the battery is 1.18 mA. When the switch is closed in position b, the current in the battery is 2.08 mA.

(a) Find the resistance R1.
= k

(b) Find the resistance R2.
= k

(c) Find the resistance R3.
= k

Please Help!!! Thank You in Advance!

Explanation / Answer

Let x be the current flowing when switch in original position
5=1(R1+R2+R3).
R1+R2+R3=5.............1.
When switch in position a R2, R2 resistors are in parallel.
Net Resistance =R2*R2/(R2+R2)=R2/2.
In this position R2/2, R1,R3 are in series with supply voltage.
5=y*(R1+R2/2+R3).
5=1.18(2R1+R2+2R3)/2.
2R1+R2+R3=10/1.18
2R1+R2+2R3=8.47.................2.
When switch in position b R3 gets shorted and by passed and R1,R2 are in series.

5=2.08*(R1+R2).
R1+R2=2.4 .................3.
Solving these three equations for R1,R2,R3 we have
R1=0.87KOhms.

R2=1.53 Kohms.

R3=2.6Kohms.

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