Question

Three 3 Ω resistors P, Q and R are connected as shown in the figure. Each of them dissipates energy and can withstand a maximum power of 27 W without melting. Find the maximum current that flows through the resistor P.

Answer 1

Answer:

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Explanation:

V = voltage across a resistor

I = current through the resistor

R = value of resistance.

Power  dissipated by the resistor is given by : P =  V²/R = I²

                                                                            I² = P / R

Each resistance dissipates energy = 27 Watts

Maximum current through Q or R or P is given by : I² = 27 Watts/3 ohms

So I = 9 Amperes maximum

Since the resistances Q and R are connected in parallel, the current that comes from the resistance P is divided equally between the resistance Q and R.

so current through resistance Q = current through resistance R = 1/2 * current through P

Since, the current passing through resistor P is max = 3 ampere, then the current through resistor Q and resistor R are 1.5 Amperes maximum.

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Answer 2

Answer:

4 Ampere

Explanation:

Total resistance=

For resistor R and Q, total resistance inverse of 1/3+1/3 which is 3/2 or 1.5ohm

For P resistance is 3 ohm

so total resistance = 4.5 ohm

Now, For each resistor

we know P=VI

or P=V²/R

then V=√(PR)

putting value......

V=√(27x3) = √81

V=9, This is voltage for each resistor in this circuit,

For R and Q we know that because they are in parallel total voltage is equally to each voltage (V=V1=V2), so we will take it for once that is 9 and for

resistor P, V=9

As we have combined R and Q therefore there voltage are in series that is 9+9=18

Total voltage =18V

Now,

V=IR

I=V/R

Putting value......

I=18/4.5

I=4 AMPERE