Question

A fax machine uses 0.106 A of current in its normal mode of operation, but only 0.040 A in standby mode. The machine uses a potential difference of 120 V. (a) In one minute, how much more charge passes through the machine in normal mode versus the standby mode, and (b) how much more energy is used?

Answer 1

In one minute, the fax machine passes 3.96 C more charge in normal mode than in standby mode. The energy used is 572.4 J more in normal mode than in standby mode.

Step-by-step explanation:

To find the charge that passes through the fax machine, we can use the equation Q = It, where Q is the charge, I is the current, and t is the time in seconds. In one minute (60 seconds), the charge that passes through the machine in normal mode is Q = 0.106 A * 60 s = 6.36 C. In standby mode, the charge is Q = 0.040 A * 60 s = 2.4 C. Therefore, the difference in charge is 6.36 C - 2.4 C = 3.96 C.

The energy used is given by the equation E = Pt, where E is the energy, P is the power, and t is the time in seconds. In normal mode, the power used is P = VI = 120 V * 0.106 A = 12.72 W. In standby mode, the power used is P = VI = 120 V * 0.040 A = 4.8 W. Therefore, the difference in energy is 12.72 W * 60 s - 4.8 W * 60 s = 572.4 J.

Answer 2

Step-by-step explanation:

Given that,

Fax machine uses a current of 0.106A on normal mode

I = 0.106A

The fax machine uses a current of 0.04A on stand by mode

I' = 0.04A

The potential different is 240V

V = 120V

A. In one minute, charges that pass through the fax machine between the standby mode and normal mode

t = 1minute = 60seconds

Charge is given

Q = it

So, the charge that pass through the fax machine is

Q = It — I't

Q = (I — i') t

Q = (0.106 — 0.04) × 60

Q = 3.96C

B. Work in electric field is given as

W = QV

Where

W = work

Q = charge

V = potential difference

W = 3.96 × 120

W = 475.2 J