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  • Three 2 omega resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors?

Three 2 Ω resistors, A, B, and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors. 

Answers (1)

Explanation:-

Given:

Maximum Power a Resistor can handle (P): 18\omega

Resistance Value of A: 2 \Omega

Resistance Value of B: 2 \Omega

Resistance Value of C: 2 \Omega
P=I^{2}R\\ I^{2}=\frac{P}{R}\\ I=\sqrt{\frac{P}{R}}\\ I=\sqrt{\frac{18}{2}}=\sqrt{9}=3A

After substituting the values in the above equation 1 we get maximum current a resistor can take is 3 amperes.

IA=IB+Ic …(2)

So, if you note from equation 2, whatever current passes through resistor A gets split and passes through resistors B and C.

Therefore, Resistor A has the maximum current of 3A passing through it due to an upper limit of power of 18W.

Hence, IA=3A 

As the resistors B and C are parallel to each other, the voltage across them is going to be the same. Let us assume it to be V
I_{B}=\frac{V}{2}\\ I_{C}=\frac{V}{2}\\
So we can see that current in Resistors B and C are going to be same and from equation 2 we can see that sum of current in both resistors is 3A. 

I_{B}=\frac{3}{2}=1.5A\\ I_{C}=\frac{3}{2}=1.5A\\

So, the maximum current that can flow through each resistor is

IA=3A

IB=1.5A

IC=1.5 A

Posted by

infoexpert24

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