In the circuit below, all four resistors are identical (R1 = R2 = R3 = R4 = R) a

Question

In the circuit below, all four resistors are identical (R1 = R2 = R3 = R4 = R) and the battery has a voltage of 5.12 V. (a) When the switch is placed in position 1, the measured current in the battery is 1.22 mA. What is the value of each resistor? k (b) When the switch is placed in position 2, what is the current flowing out of the battery? mA (c) When the switch is open (neither in position 1 or position 2), what is the current flowing out of the battery? mA

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Explanation / Answer

1) when switch is at position 1

so in this case R2 will be short circuit so

net Resistance( R1 and R4 will be in serise ) = R1+R4

= 2R

so current from batary = i = V/2R

1.22 *10-3 = 5.12 /2R

R =2098.4 ohm

2) when switch is at 2 so R3 and R4 will be in parallel and this will be in serise with R1 and R2

net resistance = R + R + R/2

= 5R/2

= 5 * 2098.4 /2

= 5245.9 ohm

so current i' = 5.12 /5245.9

=0.97 mA                                                   ans2

3) when switch is open then R3 will not be in use and R2 R1 and R4 will be in serise

eq resistance = R1+R2+R3

= 3R

=3 *2098.4

=6295.2 ohm

so current i = V/3R

i = 5.12/6295.2

= 0.81 mA                               .......    answer3

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