Discussion :: Parallel Circuits  General Questions (Q.No.15)
15.  The current through R1 will be: 

Answer: Option C Explanation: No answer description available for this question.

Seshendra said: (Nov 11, 2010)  
Voltage across parallel resistors are same. So voltage across R1 is 48v, We Know i=v/r. =48mA. 
Niraj Kumar said: (Jan 12, 2011)  
Total resistance Rt=(1*2*3)/(1+2+3)=1 I=V/Rt I=48/1*10^3=48mA 
Roopa.R said: (Sep 6, 2011)  
0.0880.040=0.048or 48mA (I1=I2+I3) SO I3=I1I2 
Vasu said: (Sep 21, 2011)  
In the answer given by Nirak kumar , Req= 1 but, its wrong. The correct formula in case of 3 r's in  is Req=R1R2R3/(R1R2+R2R3+R3R1). So here Req will become 6/11. 
Chaitanya said: (Sep 22, 2011)  
Vasu is correct 
Ragi G R said: (Jan 11, 2012)  
By simple logic we can find it. The current in the previous branch is 0.088. This current get split into two branches. One of them is 0.040. So the current in the required branch is 0.0880.040=48mA 
Yamini said: (Feb 19, 2012)  
The current in the first loop is 88mA,in the second loop is 40mA, So the current across r1 is the difference between the 2 currents,that is 88m40m=48mA 
Dipika said: (Sep 1, 2012)  
@Niraj is correct answer becoz this three resistsnce are parellel so formula is r1*r2*r3/r1+r2+r3... so total resi r.. & then after ohms law apply... 
Harika said: (Sep 4, 2012)  
@according to vasu as req=6/11 I=V/Req =48*(11/6)=88mA 
Shiva said: (Dec 13, 2013)  
I = 0.088. I1 = ? I2 = 0.040. Since I = I1+I2 (Kirchhoff current law). I1 = II2 = 0.088  0.040 = 0.048 = 48 x 10^3 = 48mA. 
Harish Kumar said: (Jan 27, 2014)  
We know voltage across parallel circuits are same. Given that V = 48v. By ohm's law, I=V/R.........(1). Total resistance R1.R2.R3/R1+R2+R3. = 1.2.3/1+2+3. = 1 Ohm. Substitute R value in 1, I = 48mA. 
Kamal said: (Jan 30, 2014)  
@Harish Kumar is wrong because the total resistance calculation is R1.R2.R3/R1R2+R2R3+R3R1. 
Prasanna Kumar said: (May 10, 2014)  
Here one simple formula is there, In parallel circuit I = I1+I2. IN THE CKT SHOWN ONE CURRENT IS LEAVING(0.088A) AND OTHER ENTERING(0.040A) THEN, I+0.040 = 0.088. I = 0.0880.040. = 88*10^3  40*10^3 = 48*10^3 = 48mA. 
Azhagusurya said: (Feb 20, 2015)  
Consider th first node: Ii = I1+I2. Ii = 0.088. & I2 = 0.040. So I1 = IiI2 = 0.0880.040. I1 = 0.048 in mA. I1 = 48mA. 
Armandwish said: (Sep 11, 2015)  
According to KCL: I = I1+I2....I1 = II2. 
Vineeth K said: (Feb 8, 2017)  
48mA is the correct answer. The total current I eff is V/Reff. Reff is equivalent resistances of r1 r2 and r3 which is 6/11. So I=48/6 * 11= 88mA. 88mA is what comes from the source which gets divided into two parts, I1 into resistor r1 and I2 into resistor r2. Which is given in the diagram through ammeter reading. V is 48v across all the resistors as the voltage does not get divided in parallel connections. So, performing I=v/r for each branch r1 r2 and r3 we get a total of 88mA with current through r1 as 48mA, r2 as 24mA and r3 as 16mA. 
Sushil said: (Dec 10, 2017)  
Thanks @Seshendra. 
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