physics-class-12-unit-03-chapter-05-electrical-energy-and-power-l-5-11-gaueyvpuba4

### <Success style="font-size:25px"> Electrical Energy And Power L-5 </Success>
### <primary style="font-size:24px"> Ohm's Law </primary>
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- $\varepsilon=\frac{d w}{d q}  =\frac{\text { Joule }}{\text { coulomb }}=\text { volt.} $

- $w  =\int \vec{F} \cdot \overrightarrow{d l} $

- $\varepsilon  =\oint \frac{\vec{F}}{q} \cdot \overrightarrow{d L}$

- $\oint$ = contour integral

- $\varepsilon = \phi {\vec{E} \cdot \overrightarrow {d {L}}}$

- Inside the battery Force is Non - Conservative.

- Field outside is Electrostatics - Conservative.

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<footer style = "font-size: 18px">Energy and Power $\rightarrow$ EMF $\rightarrow$ <span style = "color:blue"> Ohm's Law </span> $\rightarrow$ Ohm's Law $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Electrical Energy And Power L-5 </Success>
### <primary style="font-size:24px"> Problem </primary>
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* An ideal battery (emf) provides a constant voltage across its two terminals.

* Going in direction of current, as one goes through resistance potential drop: by IR.

* In going from negative to positive $\Delta V$ mereases by $\varepsilon$.

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<footer style = "font-size: 18px">Ohm's Law $\rightarrow$ Ohm's Law $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Electrical Energy And Power L-5 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- Potential drop,From B to C by $ I R_L$,C to D by $Ir$

- Through battery potential rises by $\varepsilon$ .

- $-I R_L-I T+\varepsilon=0$

- $I=\frac{\varepsilon}{r+R_L}$

- $I=\frac{10}{17+3}=0.5 \mathrm{~A}$

- $\varepsilon = 10 \mathrm{~V} $

- T = 3 $ \Omega $

- $R_L  = 17 \Omega$

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<footer style = "font-size: 18px">Ohm's Law $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Problem $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Electrical Energy And Power L-5 </Success>
### <primary style="font-size:24px"> Power Delivered by the Circuit </primary>
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- $P_{e m f}=\frac{d w}{d t}=\frac{\epsilon \cdot d \underline{a}}{d t}=\varepsilon I $

- **Joule Heat Loss**

- Amount of energy resistance absorbs and dissipates.

- $V_A>V_B$

- P.E. of d q at A.

- $V_A$ dq

- $A+B: v_B d q$

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<footer style = "font-size: 18px">Problem $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Power Delivered by the Circuit </span> $\rightarrow$ Power Delivered by the Circuit $\rightarrow$ Problem </footer>

### <Success style="font-size:25px"> Electrical Energy And Power L-5 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- $ =I^2 R \text { wires } $

- $ =\frac{P^2}{V^2} \cdot R$

- **Example:**

- Absorbs 80 watts of power from source.

- $R=\rho \cdot \frac{L}{\pi r^2}$

- $R^{\prime}=16 R $

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

- $\frac{v^2}{R^{\prime}}=\frac{V^2}{16 R}=\frac{80}{16}=5 \mathrm{~W}$

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<footer style = "font-size: 18px">Power Delivered by the Circuit $\rightarrow$ Power Delivered by the Circuit $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Power Loss in Transmission $\rightarrow$ Thank You </footer>

### <Success style="font-size:25px"> Electrical Energy And Power L-5 </Success>
### <primary style="font-size:24px"> Power Loss in Transmission </primary>
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- $ \frac{V}{R^{\prime}}=\frac{16}{16 R}=\frac{I}{16} $

- $I^2 R^{\prime}=\frac{I^2}{256} \times 16 R=\frac{I^2 R}{16}$

- Is power V.I.

- $\frac{V^2}{R^2}$ ?

- $ I^{2} R$

- $I  =\frac{100}{20}$

- $ =5 \mathrm{~A}$

- Power delivered to cct $=100 \times 5$ $=500 \mathrm{~W}$.

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<footer style = "font-size: 18px">Power Delivered by the Circuit $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Power Loss in Transmission </span> $\rightarrow$ Thank You $\rightarrow$  </footer>