Current

Add a ground to the circuit and you will see it is 1.68A.

Can you please give the full solution?

24

You can find it easily with KVL and KCL

How can I find this by thevnin's theory...

Thevenin's won't help with this problem. It is used for finding input and output of a system, not really for individual pieces in a system.

Hi @deep5712. Do you especially want to use Thevenin's or are you just looking for a solution?

I'm guessing you do want Thevenin solution. I've worked it out and will put it on here tomorrow. It's nearly midnight here now and I must get up early.

Thought I’d do it now after all. What the hell! I’m going to assume you know nothing about Thevenin’s Theorem. Using Thevenin’s you can reduce a circuit that has a load – in this case your load is the 24 Ω resistor – to simply a voltage source and a series resistance. You load can then be applied in series with this source and resistance. The advantage of doing this is that, if you wish, you can change your load and easily see the difference it makes to the complete circuit (represented by the voltage source and series resistance). In your case you simply want to calculate the current through the 24 Ω resistor – that’s fine. 1st stage is to calculate the Thevenin voltage source value. To do this, remove the 24 Ω resistor from the circuit. Now, taking the top pair of resistors, the voltage across the pair is (of course) 220V. Across the 30 Ω we have 220 x (30/50) = 132V. Looking at the lower pair, across the 50 Ω we have 220 x (50/55) = 200V. From this we can see the voltage presented across the terminals where the 24 Ω resistor was would be 200 – 132 = 68V. This is our Thevenin voltage source value. To calculate the Thevenin resistance we must ‘look into’ the place where the 24 Ω resistor was fitted. Can you see, we have a 20 Ω and 30 Ω resistor in parallel, and that combination is in series with a parallel combination of 5 Ω and 50 Ω? The 20 Ω and 30 Ω in parallel give us (20x30)/(20+30) = 12 Ω. The 5 Ω and 50 Ω in parallel give us (5x50)/(5+50) = 4.5454. Thus, our Thevenin equivalent resistance is 12 + 4.5454 Ω = 16.5454. We now have our final circuit, a voltage source of 68V in series with Thevenin equivalent resistor of 16.5454 and our ‘load’ resistor of 24 Ω. Finally, current flowing in circuit – and, therefore, through the 24 Ω resistor – is I=V/R which is 68/(16.5454 + 24) = 1.677 A. I hope this helps. If you have any questions I’m happy to help. If you want to connect a ground to the circuit as, I mentioned before, you will see the current through the 24 Ω resistor is 1.68 A.

I posted just now both circuits together. Search on my name.

Connect earth to your circuit.

Wow! Hurz has no comment on this one. What a surprise, not!