00:01
Here we will solve this problem by using nodal analysis method.
00:04
In the circuit, after interchanging the position of this battery and this register, we obtain this following circuit.
00:11
Now in the circuit, let's say that this is the node number 1 and this is the node number 2.
00:19
We take node number 2 as a reference node, that means voltage of node number 2 is equal to 0.
00:26
Now after applying nodal analysis at node number 1, we obtain this following equation.
00:31
V1 minus v2 over r1 plus v1 minus v2 over r2 plus v1 minus v2 over r2 plus v1 minus v2 over r3 plus v1 minus v2 over r4 is equal to 0.
00:50
That means algebraic sum of all outgoing currents from node number 1 is equal to 0.
00:57
From this we obtained v1 1 over r1 plus 1 over r1 plus 1 over r2 plus 1 over r3 plus 1 over r4 is equal to v over r1.
01:11
After putting all the given values here, this will become 1 over r1 is equal to 5 .08 5 .08 plus r2 10 .16, r3 25 .4, r3 25 .4.
01:47
15 .24.
01:56
V is equal to 12 volt over r1 r1 is equal to 5 .08.
02:06
From this we obtained v1 into 0 .4 -0026 12 over 5 .08.
02:18
By simplifying this we obtained v1 is equal to 5 .9016 volt.
02:34
Now let's say that current flowing through r1 is equal to i1, current flowing through r2 is equal to i2, current flowing through r3 is equal to i3 and current flowing through r4 is equal to i4.
02:48
The voltage of node 1 is equal to 5 .9016 volts...