3.21 Determine the current drawn from a 12V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.
3.21 Let the equivalent resistance of the given circuit be R’. The equivalent resistance of an infinite network is given by
R’ = 2 + or R’ = or + R’ = 2R’ + 2 + R’
R’ = = 1
Since R’ cannot be negative, hence R’ = 1+ = 2.73 Ω
Internal resistance, r = 0.5Ω
Total resistance = 2.73 + 0.5 = 3.23 Ω
Current drawn from the source = A= 3.72 A
<p><strong>3.21 </strong>Let the equivalent resistance of the given circuit be R’. The equivalent resistance of an infinite network is given by</p><p>R’ = 2 + <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>R</mi><mi>'</mi></mrow></mrow><mrow><mrow><mo>(</mo><msup><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mi>'</mi></mrow></mrow></msup><mo>+</mo><mn>1</mn><mo>)</mo></mrow></mrow></mfrac></math></span> or R’ = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn><msup><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mi>'</mi></mrow></mrow></msup><mo>+</mo><mi></mi><mn>2</mn><mo>+</mo><mi>R</mi><mi>'</mi></mrow></mrow><mrow><mrow><mo>(</mo><msup><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mi>'</mi></mrow></mrow></msup><mo>+</mo><mn>1</mn><mo>)</mo></mrow></mrow></mfrac></math></span> or <span title="Click to copy mathml"><math><msup><mrow><mrow><mi>R</mi><mi>'</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span> + R’ = 2R’ + 2 + R’</p><p><span title="Click to copy mathml"><math><msup><mrow><mrow><mi>R</mi><mi>'</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>-</mo><mi></mi><mn>2</mn><msup><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mi>'</mi></mrow></mrow></msup><mo>-</mo><mn>2</mn><mo>=</mo><mn>0</mn></math></span></p><p>R’ = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn><mo>±</mo><msqrt><mrow><mn>4</mn><mo>+</mo><mn>8</mn></mrow></msqrt></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></math></span> = 1 <span title="Click to copy mathml"><math><mo>±</mo><mo>√</mo><mn>3</mn></math></span></p><p>Since R’ cannot be negative, hence R’ = 1+ <span title="Click to copy mathml"><math><mo>√</mo><mn>3</mn></math></span> = 2.73 Ω</p><p>Internal resistance, r = 0.5Ω</p><p>Total resistance = 2.73 + 0.5 = 3.23 Ω</p><p>Current drawn from the source = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>12</mn></mrow></mrow><mrow><mrow><mn>3.23</mn></mrow></mrow></mfrac></math></span> A= 3.72 A</p>
According to this chapter, a galvanometer is used to find and measure the small electric currents in a circuit. The principle that works in a galvanometer is the electromagnetic induction.
There are two types of electricity - Static and Current electricity. The electric charges buildup on a material's surface is called the static electricity. The continuous flow of electric charge is termed as the current electricity. Current electricity is of two types - Alternating Current (AC) and Direct Current (DC). In AC, the charge direction reverses periodically and in DC, charge flows in one direction.
In simple words, current electricity can be defined as the electric charge continuously moving from one place to another along a pathway. It is measured in amperes (A). Electric current is needed for electrical devices to work.
No, in fact, it is one of the easiest chapter of class 12 Physics. Other chapters which are considered comparatively easy are Ray Optics and Electric Charges and Fields.
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