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Strike441 [17]
2 years ago
8

If a circuit has a power of 50 W with a current of 4.5 A, what is the resistance in the circuit?

Physics
1 answer:
lawyer [7]2 years ago
8 0

Answer:

{ \rm{power, \: p = current \times p.d}} \\ { \rm{50 = 4.5 \times (current \times resistance)}} \\ { \rm{50 = 4.5 \times (4.5 \times r)}} \\ { \rm{resistance =  \frac{50}{ {4.5}^{2} } }} \\  \\ { \rm{resistance = 2.5 \:  ohms}}

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7 0
3 years ago
Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou
kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

  • i = current flow = 3 A
  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

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4 years ago
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3 years ago
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