physics-class-12-unit-03-chapter-11-kirchhoffs-laws-current-and-electricity-l-9-11-06hw7xzomx8

### <Success style="font-size:25px"> Kirchhoffs Laws Current And Electricity L-9 </Success>
### <primary style="font-size:24px"> Junction Rule </primary>
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- Let us assume current coming in is positive.

- Junction A:

- Negative current going out

- $I_1+4-3+2=0$

- $I_1=-3 A$

- $I_2+1+3-2 =0$

- $I_2=-2 A$

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<footer style = "font-size: 18px">Kirchhoff's Law Current and Electricity $\rightarrow$ Kirchhoff's Law $\rightarrow$ <span style = "color:blue"> Junction Rule </span> $\rightarrow$ Voltage Rule $\rightarrow$ Example-1 </footer>

### <Success style="font-size:25px"> Kirchhoffs Laws Current And Electricity L-9 </Success>
### <primary style="font-size:24px"> Voltage Rule </primary>
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- For a static field

- $\oint \vec E . \vec dl$ = 0

- Moving in the direction of current

- $\Delta V$  = negative

- $ =I R \text { (Drop) }$

- Across a battery $\Delta V$ is positive in going from negative positive terminal.

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<footer style = "font-size: 18px">Kirchhoff's Law $\rightarrow$ Junction Rule $\rightarrow$ <span style = "color:blue"> Voltage Rule </span> $\rightarrow$ Example-1 $\rightarrow$ Example-2 </footer>

### <Success style="font-size:25px"> Kirchhoffs Laws Current And Electricity L-9 </Success>
### <primary style="font-size:24px"> Example-1 </primary>
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- +8 - $V_1$ -8-8 +6+10 = 0

- $V_1$ = 8v

- -I.1 + I.3 -4 -12 = 0

- 4I = 8

- I = 2A

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<footer style = "font-size: 18px">Junction Rule $\rightarrow$ Voltage Rule $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Example-2 $\rightarrow$ Example-2 </footer>

### <Success style="font-size:25px"> Kirchhoffs Laws Current And Electricity L-9 </Success>
### <primary style="font-size:24px"> Example-2 </primary>
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- $I_1$ and $I_2$ are 2 unknowns.

- $I_9=I_1-I_2 $

- $-2\left(I_1-I_2\right)-2 I_1+12=0 $

- $-6-2 I_2+2\left(I_1-I_2\right)=0$

- 2 Equation.

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<footer style = "font-size: 18px">Voltage Rule $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Example-2 $\rightarrow$ Example-3 </footer>

### <Success style="font-size:25px"> Kirchhoffs Laws Current And Electricity L-9 </Success>
### <primary style="font-size:24px"> Example-2 </primary>
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- $ I_1=I_2+I^{\prime \prime}-\text { (1) } $

- $3I_1 - 3I +12 = 0$

- $ I_1+I^{\prime \prime} = 4-\text { (2) } $

- $ I_1=\frac{10}{3} \mathrm{A}, I_2=\frac{8}{3} \mathrm{A}$

- $ I^{\prime \prime}=\frac{2}{3} A^3 $

- $3 I_2+ 3I^{\prime \prime}=6$
- $-\text { (3) }$

- $ I_3=I_4=\frac{1}{2} I^{\prime \prime} $
- $ =\frac{1}{3} A $

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<footer style = "font-size: 18px">Example-1 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Example-3 $\rightarrow$ Example-4 </footer>

### <Success style="font-size:25px"> Kirchhoffs Laws Current And Electricity L-9 </Success>
### <primary style="font-size:24px"> Example-3 </primary>
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- $ =\frac{6}{1+\frac{2 R}{2+R}} $

- $ =\frac{6}{1+4 / 4} = 3 A$

- $\therefore \quad R  = 1+\frac{2 R}{2+R .}$

- $R  =2 \Omega$

- Current through nearest $2 \Omega$ is $1.5 \mathrm{~A}$.

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<footer style = "font-size: 18px">Example-2 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Example-3 </span> $\rightarrow$ Example-4 $\rightarrow$ Example-4 </footer>

### <Success style="font-size:25px"> Kirchhoffs Laws Current And Electricity L-9 </Success>
### <primary style="font-size:24px"> Example-4 </primary>
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- No current through capacitors.

- P. D. Across A B= P;D $A^{\prime} B^{\prime}$

- $ =6 \mathrm{~V}$

- $ I_3=\frac{6}{4}=1.5 \mathrm{~A}$

- $-I_3 \cdot 4-\left(I_3-I_0\right) \times 2 $

- $+2+3-\left(I_3-I\right) \cdot 3 $

- I = 1.7A

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<footer style = "font-size: 18px">Example-2 $\rightarrow$ Example-3 $\rightarrow$ <span style = "color:blue"> Example-4 </span> $\rightarrow$ Example-4 $\rightarrow$ Example-5 </footer>

### <Success style="font-size:25px"> Kirchhoffs Laws Current And Electricity L-9 </Success>
### <primary style="font-size:24px"> Example-4 </primary>
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- $ I=1.7 \mathrm{~A} $

- $ I_3=1.5 \mathrm{~A}$

- $I-I_3=0.2 \mathrm{~A}$
	
- $\Delta V_{C D}=? \quad \Delta V_{C D}=\Delta V_{C A}$

- $V_C-2-0.2 \times 2=V_D$

- $V_C-V_D=2.4 V=$ Potential drop across the plates of capacitor.

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<footer style = "font-size: 18px">Example-3 $\rightarrow$ Example-4 $\rightarrow$ <span style = "color:blue"> Example-4 </span> $\rightarrow$ Example-5 $\rightarrow$ Example-5 </footer>

### <Success style="font-size:25px"> Kirchhoffs Laws Current And Electricity L-9 </Success>
### <primary style="font-size:24px"> Example-5 </primary>
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- Each resistance =R
 
- Symmetry with respect to diagonal  A D 
 
- Loop ABCD XVYA

- $\frac{5}{2} I R=V $

- $ I=\frac{2}{5} .\frac{V}{R}$

- $R=1 \Omega $

- $V=10 \mathrm{~V}$, I = 4A

- Current supplied by the battery is
- 3I= $\frac{6 V}{9 R} $

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<footer style = "font-size: 18px">Example-4 $\rightarrow$ Example-4 $\rightarrow$ <span style = "color:blue"> Example-5 </span> $\rightarrow$ Example-5 $\rightarrow$ Thank You </footer>