Question

draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 4 ohm in series with a combination of two resistors (8 ohm each) in parallel and a voltmeter across parallel combination. Each of them dissipate maximum energy and can withstand a maximum power of 16 W without melting. Find the maximum current that can flow through the three resistors.

Answer 1

Solution is in the following image hope it helps you plz mark brainliest...

Answer 2

Answer:

The current through the three resistors is 2A, 1A, and 1A each.

Explanation:

Here we have been mentioned to draw the diagram of the particular circuit having a cell, an ammeter, a key, and a resistor of 4 ohm which is in series with a combination of two resistors that are 8 ohms each. It also has a voltmeter across the parallel combination.

Now it is mentioned that each one of them dissipates the maximum amount of energy and also can withstand the maximum power of 16 W without any melting.

We have to find the maximum current that will flow through the three resistors of this particular circuit.

The diagram has been attached below.

Here we have,

 $R_{1}$= 4 Ω

$R_{2} =R_{3}$ = 8 Ω respectively.

P = 16 W

So for the resistor$R_{1}$ , the current (I) is found as follows:

$P = I^{2} R_{1}$

⇒ 16 W = $4 I^{2}$

⇒ $I^{2} = \frac{16}{4}$

⇒ $I^{2} = 4$

⇒ $I = 2 A$

So the current across the first resistor is 2A.

Now we know that the same current will flow through the combination of the two parallel resistors as it is in series with the 4-ohm resistor and the current remains the same in series connection.

Since both resistors are equal therefore the current through each one of them is 1A each making the total to be 2A.

So the currents through the three resistors are 2A, 1A, and 1 A respectively.