Question

In the circuit shown in the figure (Figure 1) all the resistors are rated at a maximum power of 2.10 W. What is the maximum emf epsilon that the battery can have without burning up any of the resistors?

Answer 1

The current that is passing through 40 Ω resistor is equal to the emf of the circuit because the current passing through remaining resistors is less than the current through the 40 Ω resistor The maximum power in the circuit is 2.10 W The current through the 40Ω resistor is. 2.10 W-12(40 Ω) The equivalent resistance of the upper parallel combination of resistors is, P-IR I 0.23 A 20 20 A, 10 Ω The equivalent resistance of the series combination of resistors is, -10+10 20Ω Again calculate the same as above for upper branch

25 15 20 R 6.38 The parallel combination of the parallel resistors in the lower branch of the circuit is, 50 50 -252 Now all the resistors in the circuit are in series combinatiorn The equivalent resistance is, (%),-25 +30+638+40+25 = 126.38 Ω The emf of the circuit is, (0.23) (126.38)