 ×     #### Thank you for registering.

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

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

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
```
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

``` 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 & RegardsAarti Guptaaskiitians Faculty

```
6 years ago
```							Thanku mam can u give me one more example.   Of some complicated circuit
```
6 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Electric Current

View all Questions »  ### 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