**(1) A current I ampere gets divided and flows into a parallel network of resistors as shown in the adjoining figure. If the power dissipated in the 1 Ω resistor is P watt, what is the total power (in watt) dissipated in the branch containing the two 3 Ω resistors?**

**(a) P**

**(b) 6P**

**(c) 3P**

**(d) P/2**

**(e) 3P/2 **

If the current through the 1 Ω resistor is I_{1}, the current through the branch containing the two 3 Ω resistors is I_{1}/2.

[Note that the P.D across the three branches is the same and hence the current through the 6 Ω branch must be half the current through the 3 Ω branch].

The power dissipated in the 1 Ω resistor is

I_{1}^{2 }×1= P

Therefore, the power dissipated in the branch containing the *two* 3 Ω resistors is

(I_{1}/2)_{ }^{2}×6 = 6P/4 = 3P/2 watt.

**(2) A battery of internal resistance R ohm and output V volt is connected across a variable resistor. The heat generated in the variable resistor is maximum when the current in it is **

**(a) V/2R**

**(b) V/4R**

**(c) 4V/R**

**(d) V/R**

**(e) None of the above**

You may be remembering the *maximum power transfer theorem* which states that a direct current source will transfer maximum power to a load when the resistance of the load is equal to the internal resistance of the source.

In the above problem, since the internal resistance of the battery is R, the power dissipated in the variable resistor is maximum when its resistance also is R. The total resistance in the circuit in this condition is R+R = 2R and the current is V/2R.

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