00:01
Here we are solving this problem by using kirchhoff voltage low.
00:03
To apply kirchof voltage low, assume that current flowing through this circuit is equal to i.
00:11
Now by applying kirchof voltage low, we obtain this equation.
00:16
32 minus 12k into i minus 24k into i minus 12k into i minus 12k into i minus 16.
00:30
16 minus 12k into i minus 12k into i is equal to 0.
00:39
That means total voltage drop in a loop is equal to 0.
00:43
In this battery, current is leaving the positive terminal, that's why here positive sign appears.
00:48
In this battery, current is leaving the negative terminal, that's why here negative sign appears.
00:56
And you know, register always offers a voltage drop, that's why here, here, here and here negative sign appears.
01:06
32 minus 16 is equal to 16 minus 12k plus 24k plus 12k plus 12k is equal to 72k into i is equal to 0.
01:22
From this we obtain i is equal to 16 over 72 into thousand.
01:30
From this we obtain current i is equal to 2 over 2 over 1.
01:34
9mm.
01:38
Now in the circuit, voltage drop across this register is equal to vr1, voltage drop across this register is equal to vr2, voltage drop across this register is equal to vr3, voltage drop across this register is equal to vr4, voltage drop across this register is equal to vr5.
02:05
Now by using omslow, vr1 is given by vr1 is equal to current flowing through r1, that is i into r1.
02:14
Here i is equal to 2 over 9 into 10 to the power negative 3 and r1 is equal to 12 into 10 to the power 3...