00:01
So, here we have to calculate the value of the three currents in the circuit.
00:04
So, this is the circuit where we are given the value of plus vcc current that is equals to 10 volt from this interval where we are having the value of rc that is equals to 2 kilo ohm.
00:15
Here we are having the value of ib that is represented by rb which is equals to 1 mega ohm and from here the current ib is moving where we are given a pn junction diode which is given as the ic current from here and from here that is emitter current there is collector current which is represented to a 1 kilo ohm resistor which is represented by re and from here this capacitor cg is connected.
00:46
So, this is the answer basically this is basically the diagram here we have to find out the value of the three currents in the circuit.
00:54
So, this is circuit if we apply the kvl in the so we can say that vcc minus rb multiplied by the ib minus 0 minus ie multiplied by the re that is equals to 0.
01:06
So, solving the term for here so vcc minus 10 raised to the power 6 of ib minus 1 plus beta of 10 raised to the power 3 of ib that from here is equals to 0.
01:15
So, 10 is equals to ib multiplied by the 10 raised to the power 6 plus 101 multiplied by the 10 raised to the power 3...