Find currents using KVL

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Electrical Circuits, Electrical Circuits Problems, Resistive Circuits

Find resistor currents using KVL.

Solution:

$R_1$ and $V_1$ are parallel. So the voltage across $R_1$ is equal to $V_1$ . This can be also calculated using KVL in the left hand side loop:

$-V_1+V_{R_1}=0 \rightarrow V_{R_1}=V_1=10V$ .
Now, use Ohm’s law to find $I_{R_1}$ :

$V_{R_1}=R_1 \times I_{R_1} \rightarrow I_{R_1}=\frac{V_{R_1}}{R_1}=0.1 A$ .
To find $I_{R_2}$ , write KVL around the outer loop:

$-V_1+V_{R_2}-V_2=0 \rightarrow V_{R_2}=V_1+V_2=20V$ .
Again, use Ohm’s law to determine $I_{R_2}$ :

$V_{R_2}=R_2 \times I_{R_2} \rightarrow I_{R_2}=\frac{V_{R_2}}{R_2}=0.2 A$ .

Now, tell me what is the current passing through $V_1$ ?

Independent Sources KVL KVL_KCL linear Ohm’s law Source

Comments

13 responses to “Find currents using KVL”

September 30, 2013

harish

0.3

Reply
1. September 30, 2013
  
  Yaz
  
  good job!
  From KCL:
  $I_{V_1}=I_{R_1}+I_{R_2}=0.3 A$
  
  Reply
October 3, 2013

yongule

good work i got aleast some concepts

Reply
1. October 3, 2013
  
  Yaz
  
  good, very good. let me know if you need any help.
  
  Reply
  1. October 12, 2013
    
    shahidul islam
    
    Please help me to solve the current flow through the below circuit.
    
    My Circuit
    
    Reply
    1. October 12, 2013
      
      Yaz
      
      $10 \Omega$ and $3 \Omega$ are both short circuited. If you check the circuit you can easily see that both terminals are connected to each other for both.
      So, you have a circuit with a $10 V$ source and $2 \Omega$ and Ohm’s law says:
      $I=\frac{V}{R}=5 A$
      Let me know if you need more help.
      
      Reply
      1. October 14, 2013
        
        shahidul islam
        
        Thank you very much, sir.
        
        Surprisingly, in every guidebook I found the answer 7.5 Amp where there was no explanation behind the answer. But now I know the explanation and also know that correct answer is 5 Amp.
        
        I will post here if I need further help. Thanks again for your help.
      2. October 14, 2013
        
        Yaz
        
        You are welcome.
    2. May 24, 2019
      
      prasad
      
      eliminate 10ohm resister becuase current will flow through low resistance patch then it is a series resistance circuit easy you can calculate otherwise let me know
      
      Reply
October 5, 2013

Thomas

What happens if you reverse the polarity of V2?

Reply
1. October 6, 2013
  
  Yaz
  
  good question.
  Actually, it will be quite interesting. The voltage across $R_1$ is $10 V$ and therefore, the voltage drop on $R_2$ will be zero. You can also verify this by KVL around the outer loop:
  $-10 V +V_{R_2}+10 V=0 \rightarrow V_{R_2}=0$
  And using the Ohm’s law, we conclude that no current passes through $R_2$ . The current of $R_1$ is the same as before because its voltage is not changed.
  
  Reply
May 30, 2018

Min Dhami

please give me solve two voltage source are given

Reply
July 5, 2019

sony

please help me learn kvl in easy way

Reply