A current of 3.4 A flows through a battery for 2 min. How much charge passes thr

Question

A current of 3.4 A flows through a battery for 2 min. How much charge passes through the battery in that time? A current of 1.5 A is flowing through a resistance of 22 Ohm. What is the voltage difference across this resistor? Four 18-Ohm resistors are connected in series to a 9-V battery of negligible internal resistance. a) What is the current flowing through each resistor? b) What is the voltage difference across each resistor? In the circuit shown, the 2-Ohm resistance is the internal resistance of the battery and can be considered to be in series, as shown, with the battery and the 28-Ohm load. a) What is the current flowing through the 28-Ohm resistor? b) What is the voltage difference across the 28-Ohm resistor? Two resistors, each having a resistance of 30 Ohm, are connected in parallel. What is the equivalent resistance of this combination?

Explanation / Answer

1.

i = dq/dt

dq = i*dt

dt = 2 min = 120 sec

dq = 3.4*120 = 408 C

2.

ohm's law:

V = i*R

V = 1.5*22 = 33 Volt

3.

four 18 ohm resistor in series

total resistance in series circuit is given by

Req = R1 + R2 + R3 + R4

Req = 18 + 18+ 18 +18 = 72 ohm

V = 9.0 Volt

V = i*Req

i = V/Req

i = 9/72 = 1/8 = 0.125 Amp.

Now in series current is distributes equally in each resistor.

So current in each resistor will be

i1 = i2 = i3 = i4 = 0.125 Amp.

Voltage in each resistor will be same (only in this case because all the resistors are same)

V1 = i1*R1 = 0.125*18 = 2.25 V

V1 = V2 = V3 = V4 = 2.25 Volt

4.

figure is not shown.

5.

in parallel circuit

1/Req = 1/R1 + 1/R2

1/Req = 1/30 + 1/30

1/Req = 1/15

Req = 15 ohm

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