Lsn 21-Source of magnetic Fields II Problem 25.51 Constants 1 Periodic Table Par

Question

Lsn 21-Source of magnetic Fields II Problem 25.51 Constants 1 Periodic Table Part A The voltage across the 30-kf2 resistor in the figure (Figure 1) is measured with (a) a 90-k32 voitmeter, (b) a 400-k voltmeter, and (c) a digital meter with 10 Mn resistance. To two significant figures, what does each read? Express your answer using two significant figures. v. 23 Previous Answers Request Answer Figure 1011 > X Incorrect; One attempt remaining; Try Again Part B 30 k Express your answer using two significant figures. bg= 58 V 100 V 40 k 40 k Previous Answers Corret

Explanation / Answer

Solution :-

We have to find the voltage across each resistor.

We can see in the figure that the two 40 k Ohm resistors are connected in parallel so their equilavent will be:

1/R = 1/40 + 1/40

R = 40 x 40 / (40 + 40) = 1600/80 = 20 k Ohm

In order to measure the voltage across 30 kOhm, the voltmeter will be connected to it in parallel.

(a)r = 90 k Ohm

The equivalent resistance will be:

R' = 30 x r / (30 + r) = 30 x 90 / (30 + 90) = 4500 / 90 = 90 Ohm

The total resistance of the circuit will be:

Req = R + R' (As the combination of the two are in series)

Req = 20 + 30 = 50 k Ohm

From Ohm's law: V = IR => I = V/R = VReq

I = 100/50 x 103 Ohm = 2 x 10-3 A

So the voltage across 30 k Ohm will be:

V(30) = I x R' =  2 x 10-3 A x 30x 103 Ohm = 60 Volts

Hence, V(30) = 60 V

(b)r = 400 k Ohm

R' = 30 x 400 / (30 + 400) = 27.91 k Ohm

Similarly as the above part:

Req = R + R' = 20 + 27.91= 47.91 k Ohm

I = 100 / 47.91x 103 = 2.09x 10-3 A

The voltage across, 30 k Ohm in this case will be:

V(30) =  2.09 x 10-3 A x  27.91 x 103 = 58.332 V

Hence, V(30) = 58.332 V

(c)r = 10 x 106 Ohm = 10000 k ohm

R' = 30 x 10000 / (30+ 10000) = 29.91 Ohm

Req = 29.91 + 20 = 49.91 k Ohm

I = 100 / 49.91 x 103 = 2.003 x 10-3 A

drop across 30 k Ohm will be:

V(30) = 29.91 x 103 x 2.003 x 10-3 A = 59.91 Volts

Hence, V(30) = 59.91 V

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