physics-class-12-unit-03-chapter-06-series-and-parallel-combination-of-resistances-l-6-11-n-qjzm2v3ek

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Potential </primary>
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* $\varepsilon I$

* Potential Drop along a circuit

- $V_A=V_B+I R$

- **3.** $ I^2 R=\frac{V^2}{R}$

- V= Potential difference across in the ends of resistor(Ohmic).

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<footer style = "font-size: 18px"> $\rightarrow$ Series and Parellel Combinations of Resistances $\rightarrow$ <span style = "color:blue"> Potential </span> $\rightarrow$ Power $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Power </primary>
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- P = Power to be delivered.

- $R_c=$ Resistance of cables.

- $P_c=I^2 R_c=\frac{P^2}{{v}^2} \cdot R_c \quad$

- $P=I \cdot V$

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<footer style = "font-size: 18px">Series and Parellel Combinations of Resistances $\rightarrow$ Potential $\rightarrow$ <span style = "color:blue"> Power </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Example </primary>
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- The section AC absorbs 64 watts of power.

- $ V_{A C}=\frac{64}{I}=\frac{64}{1.6}=40 \mathrm{~V} $

- $ V_{A B}=I R=1.6 \times 20=32 \mathrm{~V}$

- Voltage provide by battery $=40-32=8 \mathrm{~V}$ Resistance absorbs

- $I^2 R=(1.6)^2 \times 20=2.56 \times 20=51.2 \text { watts. }$

- Power absorbed by the battery

- 64 - 51.2 = 12.8 $\mathrm{~W}$

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![image](https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/physics-class-12-unit-03-chapter-06-series-and-parallel-combination-of-resistances-l-6_11-n_qjzm2v3ek-03.jpg)

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

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Example </primary>
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- 100 Watts. -switched on for 30 days.
- Household unit of electricity Rs 3/-
- 220-240 V

- $\frac{V^2}{R}=100$

- $ \frac{57600}{100}=R \Rightarrow 576 \Omega $

- I V  = 100

- $I  =\frac{100}{V}=\frac{100}{240}=\frac{5}{12} \mathrm{~A}$

- 24 units per day.72 units (KWH) per month.

-  Rs216 /-

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

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Example </primary>
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- Two Bulbs 60 W , 90 W to be put in $240 \mathrm{~V}$ mains.

- Series combination.

- $ P_1=\frac{V^2}{R_1} \Rightarrow 60=\frac{\left(240 R^2\right)}{R_1}$

- $R_1=\frac{57600}{60}=960 \Omega$

- $ P_2=\frac{V^2}{R_2}=90 \Rightarrow R_2=640 \Omega $

- $ V=I R_1+I R_2 \Rightarrow I=\frac{V}{R_1+R_2}=\frac{240}{1600}=\frac{3}{20} 4 $

- $ V_1=\frac{3}{20} \times R_1=\frac{3}{20} \times 960=144^{\circ} \mathrm{V}$

- $ V_2=\frac{3}{20} \times R_2=96 \mathrm{~V}$

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Resistances in Parallel </footer>

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Example </primary>
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- $P_1  =I^2 \cdot R_1 $
- $ =\frac{9}{400} \times 960=21.6 \mathrm{~W}$

- $P_2  =14.4 \mathrm{~W}$

- Total power $=36 \mathrm{~W}$

- **Parallel Combination**

- Voltage across the resistances are this same.

- Total Power $=150 \mathrm{~W}$

- $I_1=\frac{240}{960}=\frac{1}{4} \mathrm{~A} $

- $I_2=\frac{240}{640}=\frac{3}{8} \mathrm{~A} $

- $P_1=I_1^2 R_1=\frac{1}{16} \times 960=60 \mathrm{~W}$

- $P_2=I_2^2 R_2=\frac{9}{64} \times 640=90 \mathrm{~W}$

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Resistances in Parallel $\rightarrow$ Equivalent Resistance </footer>

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Resistances in Parallel </primary>
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- $I=I_1+I_2$

- When a potential difference is applied across the combination ( $a$ to $b$ ) the same P.D. appears across each branch.

- A single resistance which can replace the combination such that I remains same.

- $I_1=\frac{\Delta V}{R_1} \quad I_2=\frac{\Delta V}{R_2} \text {. }$

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Resistances in Parallel </span> $\rightarrow$ Equivalent Resistance $\rightarrow$ Series Combination </footer>

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Equivalent Resistance </primary>
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- $I  =I_1+I_2 $
- $ =\frac{\Delta V}{R_1}+\frac{\Delta V}{R_2}$
- =$\Delta V\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$

- if $R_{\text {eq }}=$ Equivalent resistance.

- $I=\frac{\Delta V}{R_{e q}}$

- $\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}$

- $\frac{1}{R_{e q}}=\sum_n \frac{1}{R_n}$

- Req is smaller than the smallest resistance in the circuit.

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<footer style = "font-size: 18px">Example $\rightarrow$ Resistances in Parallel $\rightarrow$ <span style = "color:blue"> Equivalent Resistance </span> $\rightarrow$ Series Combination $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Series Combination </primary>
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- A Same current through each member.

- $\Delta V  =\Delta V_1+\Delta V_2 $

- $ =I R_1+I R_2 $

- = $I\ (R_{1}+R_{2}) $

- $R_{e q}  =\sum_i R_i$

- Series NOT Pananel.

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<footer style = "font-size: 18px">Resistances in Parallel $\rightarrow$ Equivalent Resistance $\rightarrow$ <span style = "color:blue"> Series Combination </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Example </primary>
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- $ V_{a b}=I R_1 $

- $ V_{b c}=I R_{e q}= I \cdot \frac{R_2 R_3}{R_2+R_3}$

- $V_{a c}  =V_{a b}+V_{b c} $

- = $ I R_1+I \frac{R_2 R_3}{R_2+R_3} $

- $I  =\frac{V_{a c}}{R_1+\frac{R_2 R_3}{R_2+R_3}}$

- =$\frac {V_{a c} (R_2 + R_3)}{R_1 R_2+R_2 R_3+R_3 R_1}$

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<footer style = "font-size: 18px">Equivalent Resistance $\rightarrow$ Series Combination $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Example </primary>
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- $R_2 \|| R_3$

- Equivalent Resistance

- $R_{e q}  =\frac{R_2 R_3}{R_2+R_3} $
- $ =\frac{2 \times 3}{2+3}=\frac{6}{3} \Omega$

- $R_{1}$ in series will $R_{eq}$

- New equivalent resistance

- $2+\frac{6}{5}=\frac{16}{5} \Omega$

- $R_1 = 2\Omega$

- $R_3 = 3\Omega$

- $R_{eq} = \frac{6}{5}$

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

### <Success style="font-size:25px"> Series And Parallel Combination Of Resistances L-6 </Success>
### <primary style="font-size:24px"> Example </primary>
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- Before switch is closed.

- $R_{eq} = \frac {20 \times 12}{20 + 12} = 7.5{\Omega}$

- I = $\frac {21}{7.5}$ = 2.8A

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