badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade:

                        

How to find out the current through a circuit having resistances in series and parallel and having only on battery using kirchoff`s law

6 years ago

Answers : (2)

Aarti Gupta
askIITians Faculty
300 Points
							
Series and parallel circuits are complex circuits.In other words, there will be parts with different voltages and parts with different currents. The rules for solving these circuits are the same , just apply the series rules for the series components and parallel rules for parallel components. These circuits can be very complex with many different voltage divisions and current divisions.
Kirchhoff's Current Law can be stated as "the sum of the currents entering a point must equal the sum of the currents leaving that point".Expression can be IT = I1 + I2 + . . . + IN

In the original circuit, on top , it can be seen that current flows from the negative end of the source (the bottom) counterclockwise around the circuit. It would flow up through R3 and then divide into I1 through R1 and I2 through R2, and then recombine into IT at the top before returning to the source. This makes R1 and R2 in parallel with each other. The total current (IT) and I3 would be the same and according to Kirchhoff's Current Law, I1 + I2 = IT. So this circuit has three different current values. Note since R1 and R2 are in parallel they would have the same voltage, but different currents.

To solve first find the equivalent resistance of R1 and R2 in parallel. Req = 1/(1/1200 + 1/2400) = 800 W. Now we can redraw the circuit and see that this R1-R2 equivalent is in series with R3. RT then is the sum of these two values. RT = 800 + 1200 = 2000 W. IT = V/RT = 12/2000 = 0.006 A or 6 mA.

The simplified circuit shows that there will be two voltage drops, one across the parallel combination and one across R3. These can be found as Veq = IReq = 0.006 x 800 = 4.8 V and V3 = IR3 = 0.006 x 1200 = 7.2 V. Kirchhoff's Voltage Law says that the two drops should add to equal VS and 4.8 V + 7.2 V = 12 V. So far so good.

Now that we know that V1 and V2 = 4.8 V, I1 and I2 can be found. I1 = V1/R1 = 4.8/1200 = 0.004 A or 4 mA, and I2 = V2/R2 = 4.8/2400 = 0.002 or 2 mA. Kirchhoff's Current Law says that these two currents should add to equal the current that they split from and 2 mA + 4 mA = 6 mA.

Thanks & Regards

Aarti Gupta

askiitians Faculty

6 years ago
Priya Bahuguna
8 Points
							Thanku mam can u give me one more example.   Of some complicated circuit
						
6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details