Question

When the switches are closed in circuit 2,ammeter 1 read 6A and ammeter 2

reads 3A. Calculate

i) the current in resistor P,

ii) the power supplied to resistor Q and

iii) the energy transformed in resistor Q in 300 sec​

Answer 1

Answer:

Explanation:Current flowing through Q ia 3A.

Hence, power dissipated Pq = 3* 100= 300 Watts.

Power supplied Q in 300 seconds is (300/3600)*300= 25 Watts

Current through P is 6A-3A = 3A

Answer 2

(I). The current in resistor P is 3 A.

(II). The power supplied to resistor Q is 30 Watt.

(III). the energy transformed in resistor Q in 300 sec​ is 9000 J.

Explanation:

Given that,

Current in 1 ammeter = 6 A

Current in 2 ammeter = 3 A

(I). We need to calculate the current in resistor P

According to Kirchhoff's law

The current of 1 ammeter is divided in two mash.

So, The current in resistor P is 3 A.

(II). We need to calculate the power supplied to resistor Q

Using formula of power

$P=I\times R$

Put the value into the formula

$P=3\times 10$

$P=30\ Watt$

(II). We need to calculate the energy transformed in resistor Q in 300 sec​

Using formula of energy

$E=P\times t$

Put the value into the formula

$E=30\times300$

$E=9000\ J$

Hence, (I). The current in resistor P is 3 A.

(II). The power supplied to resistor Q is 30 Watt.

(III). The energy transformed in resistor Q in 300 sec​ is 9000 J.

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Topic : electricity

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