Question

In the circut below, the lighbub has a resistance of 300. The top batery is 100V with an internal resistance de 10 (drawn night need to 1). The botom batery is 9.0V with internal resistance What is the convenional current through the lighbub, and what direction ist? -TA (v l.on (A) 10A right to let (B) 030A, right to let (C) 10A, le bright (D) 030A, let bright

Answer 1

Answer:- D) 0.3 A left to right

Explanations:-

Here as per the given diagram lets use nodal analysis at the node which is near to the light buld by considering the voltage at that point as Vx,

So that here there is no any internal resistance for 9V battery then the voltage across the bulb will also be 9 V

Vx = 9 V

When only considering 9 V battery then

I1 = Vx/30 = 9/30 = 0.3 A

And we consider only 10 V battery then as the 9 voltage path becomes shortened so that there will not be any current through the bulb and hence it will not given any direction to it also

Hence the total current through the bulb will be

I = 0.3 A

And its direction is to the right