Question

You are being provided a battery, a key, three resistance, an ammeter and a voltmeter. You make a circuit after connecting these with the wires

Answer 1

Answer: Here the circuit in which a resistor of resistance 4Ω is attached in series and two resistors of resistance 8Ω are attached in parallel. Also, Ammeter and voltmeter are attached to the circuit in series and parallel respectively.

Now,

The maximum power of each resistor is 16W

4Ω resistor is joined in series with the circuit.

so, Power=i

2

R , here i is current through resistor of resistance R

16=i

2

∗4Ω

i = 2A

Now, 2A is passing through parallel resistors of resistance 8Ω.

we know, in parallel potential difference be constant so,

current divides two-part, due to same resistance current on each resistance will behalf. so, the current passing through each parallel resistor is  

2A

2

​

=1A

So, finally current through 4Ω resistor =2A

current through each 8Ω resistor =1A

solutions maximum power of each resistor is 16W

4Ω resistor is joined in series with the circuit.

so, Power=i

2

R , here i is current through resistor of resistance R

16=i

2

∗4Ω

i = 2A

Now, 2A is passing through parallel resistors of resistance 8Ω.

we know, in parallel potential difference be constant so,

currently divides two-part, due to same resistance current on each resistance will behalf. so, the current passing through each parallel resistor is  

2A

2

​

=1A

So, finally current through 4Ω resistor =2A

current through each 8Ω resistor =1A

Explanation: The maximum power of each resistor is 16W

4Ω resistor is joined in series with the circuit.

so, Power=i

2

R , here i is current through resistor of resistance R

16=i

2

∗4Ω

i = 2A

Now, 2A is passing through parallel resistors of resistance 8Ω.

we know, in parallel potential difference be constant so,

currently divides two-part, due to same resistance current on each resistance will behalf. so, the current passing through each parallel resistor is  

2A

2

​

=1A

So, finally current through 4Ω resistor =2A

current through each 8Ω resistor =1A

Answer 2

Answer:

Here the circuit in which a resistor of resistance 4Ω is attached in series and two resistors of resistance 8Ω are attached in parallel. Also, Ammeter and voltmeter attached in the circuit in series and parallel respectively.

Now,

Maximum power of each resistor is 16W

4Ω resistor is joined in series with circuit .

so, Power=i

2

R , here i is current through resistor of resistance R

16=i

2

∗4Ω

i = 2A

Now, 2A is passing through parallel resistors of resistance 8Ω.

we know, in parallel potential difference be constant so,

current divides two part , due to same resistance current on each resistance will be half . so, current passing through each parallel resistor is

2A

2

=1A

So, finally current through 4Ω resistor =2A

current through each 8Ω resistor =1A