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katovenus [111]
3 years ago
7

The rechargeable batteries for a laptop computer need a much smaller voltage than what a wall socket provides. Therefore, a tran

sformer is plugged into the wall socket and produces the necessary voltage for charging the batteries. The batteries are rated at 9.0 V, and a current of 268 mA is used to charge them. The wall socket provides a voltage of 120 V. (a) Determine the turns ratio of the transformer. (b) What is the current coming from the wall socket? (c) Find the average power delivered by the wall socket and the average power sent to the batteries.
Physics
1 answer:
mart [117]3 years ago
5 0

Answer:

(a) 0.075

(b)20.1 mA

(c) 2.412 W

Explanation:

Vs = 9 V, is = 268 mA = 0.268 A

Vp = 120 V

(a) Let the number of turns in primary coil is Np and the number of turns in secondary coil of the transformer is Ns.

Ns / np = Vs / Vp

Ns / Np = 9 / 120

Ns / Np = 3 : 40 = 0.075

(b) Let the current drawn from the wall socket is ip.

Ns/ Np = ip / is

0.075 = ip / 0.268

ip = 0.0201 A = 20.1 mA

(c) Power delivered by the socket = Vp x ip = 120 x 0.0201 = 2.412 W

Power sent to the batteries = Vs x is = 9 x 0.268 = 2.412 W

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3 years ago
Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and
Liono4ka [1.6K]

a) 5 \Omega, 1.6 A

b) 6 \Omega, 1.33 A

Explanation:

a)

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

R=R_1+R_2

where

R_1=4\Omega

R_2=1 \Omega

Therefore, the equivalent resistance is

R=4+1=5 \Omega

Now we can use Ohm's Law to find the current flowing through the circuit:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=5\Omega is the equivalent resistance of the circuit

Substituting,

I=\frac{8}{5}=1.6 A

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

b)

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

R=R_1+R_2+R_3

where:

R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega

Substituting, we find the equivalent resistance:

R=4+1+1=6 \Omega

Now we can find the current through the circuit by using again Ohm's Law:

I=\frac{V}{R}

where

V = 8 V is the voltage supplied by the battery

R=6\Omega is the equivalent resistance

Substituting,

I=\frac{8}{6}=1.33 A

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

3 0
3 years ago
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