CBSE Class 12-science Answered

Asked by my3shyll | 24 Sep, 2018, 19:33: PM

(1)
Applying to Kirchoff's voltage law,
For loop A-B-D-E-A we get,
-RI -10I + 10 = 0
→ 10 = RI +10I1 ...(1)
Applying Kirchoff's voltage law for loop F-A-B-C-F,
-2 - RI + 5 I2 = 0
→5 I2 - RI = 2 ...(2)
We know,
by Kirchoff's current law,
I1 = I + I2
→ I2 = I1 - I
Substituting it in equation (2),
5 (I1 - I) - RI = 2 ... (3)
multiplying equation (3) by '2'
10I1 - 10I- 2 IR = 4 ...(4)
Subtracting equation (4) from (1),
RI +10I1 = 10
- ] 10I1 - 10I- 2 IR = 4
(-) (+) (+)
3RI + 10 I = 6 ...(5)
Equation (5) can be written as,

By Ohm's law,
Veff= I (R + Reff)
Comparing thios equation with (6),
Veff= 2 Volt and Reff = 10/3 = 3.33 Ω
The equivalent circuit for figure (1) is,

Answered by Shiwani Sawant | 27 Sep, 2018, 13:09: PM
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