Given the circuit below. The voltage is 5 volts and each resistor is 10 ohms. Th

Question

Given the circuit below. The voltage is 5 volts and each resistor is 10 ohms. The current flowing through the circuit (when measured by an ideal ammeter) is 1 amp.

a) Show current through the circuit is 1 A.

Next, assume you put a DMM (in ammeter mode) which has a resistance of one ohm at point A. Recalculate the equivalent resistance of the circuit.

b) What is the current flowing through the circuit now. This is the current reading the DMM will give as flowing into the junction indicated by the arrow. Is it still 1 A?

Move the DMM to measure I1.

c) The resistance through the top branch is now 11 ohms (resistor + ammeter). Calculate the new current.

You now move ammeter to measure the I2 .

There is no need to redo calculation since current will now be the same as calculated for I1 since both resistors are the same.

d) Is the measured current into the (indicated) junction the same as the sum of the currents measured out of the junction?

Explanation / Answer

a)The requivalent resistance is:

Req = R1 x R2/(R1 + R2) = 10 x 10/(10 + 10) = 100/20 = 5 Ohm

from Ohm's law, V = IR => I = V/R

I = 5/5 = 1 A

Hence it is proved that the current in the circuit is 1 A

Req' = 1 + 5 = 6 Ohm

I = 5/6 = 0.833 A

Hence, I' = 0.833 A

b)V1 = V2

I1/R1 = I2/R2 =. I1 = I2

but I1 + I2 = 0.833 A

2I1 = 0.833 => I1 = 0.417 A

Hence, I1 = 0.417 A

c)Req = 11 x 11 / (11 + 11) = 5.5 A

I = 5/5.5 = 0.91 A

I = 1 A (aprox)

d)Yes

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