How long would it take for a net charge of 2.5 C to pass a location in a wire if it is to carry a steady current of 5.0 mA? (b) If the wire is actually connected directly to the two electrodes of a battery and the battery does 25 J of work on the charge during this time, what is the terminal voltage of the battery?

Given Information:

Charge = Q = 2.5 C

Current = I = 5 mA = 0.005 A

Work done = E = 25 J

Required Information:

time = t = ?

Voltage = V = ?

Answer:

time = 500 seconds

Voltage = 10 volts

Explanation:

As we know the current flowing in a circuit is given by

I = Q/t

t = Q/I

t = 2.5/0.005

t = 500 seconds

As we know the voltage is given by

V = E/Q

V = 25/2.5

V = 10 volts

(a) It would take 500 s

(b) The terminal voltage is 20 V

Explanation:

(a) Time = charge (Q) : current (I)

Q = 2.5 C

I = 5 mA = 5*10^-3 A

Time = 2.5 : 5*10^-3 = 500 s

(b) Terminal voltage = 2*workdone/charge = 2*25/2.5 = 20 V