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SS1: PHYSICS – 3RD TERM

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  1. Production of Electric Current | Week 1
    6 Topics
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    1 Quiz
  2. Electric Current | Week 2
    5 Topics
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    1 Quiz
  3. Electrical Resistance of a Conductor | Week 3
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    1 Quiz
  4. Particulate Nature of Matter | Week 4
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  5. Crystalline and Non-crystalline Substances | Week 5
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  6. Elastic Properties of Solids | Week 6 & 7
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  7. Fluids at Rest & in Motion | Week 8 & 9
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  8. Solar Collector
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Resistance in Parallel:

When resistors are placed in parallel, they are arranged side by side with their ends joined together. With the arrangement, the potential difference across the resistors is the same.

The positive terminals of all the resistors are connected to the same positive terminal and the negative terminals of the resistors are connected to the same negative terminal.

resistors in parallel

The equipment parallel resistance Rtotal is found using the expression,

I = I1 + I2 + I3 ____________________ (1)

Since the current flowing through each resistor is 

I1 = \( \frac{V}{R_1}\), I2 = \( \frac{V}{R_2}\), I3 = \( \frac{V}{R_3}\)   __________________ (2)

Substituting these equations (2) in (1)

 ∴ I = \( \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} =\frac{V}{R_{total}}\)

= \( \scriptsize V \left ( \normalsize \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right) = \normalsize \frac{V}{R_{total}}\)

cross multiply

\(\frac{V}{V(R_{total})} = \frac{1}{R_1}+ \frac{1}{R_2} + \frac{1}{R_3}\)

∴ \(\frac{1}{R_{total}} = \frac{1}{R_1}+ \frac{1}{R_2} + \frac{1}{R_3}\)

For two resistors in parallel

Screenshot 2022 05 25 at 11.23.00

Total Resistance = \( \frac {R_1 R_2}{R_1 + R_2} \)

Example 1:

Three resistors 6 Ω, 8 Ω and 10 Ω are connected in parallel. Calculate the total resistance of the circuit.

Solution:

Data given in the question:

R1 = 6 Ω, R2 = 8 Ω, R3 = 10 Ω

Formula: \(\frac{1}{R_{total}} = \frac{1}{R_1} \: + \: \frac{1}{R_2} \: + \: \frac{1}{R_3}\)

Substitute the values into the equation:

⇒ \(\frac{1}{R_{total}} = \frac{1}{6}+ \frac{1}{8} + \frac{1}{10}\)

L.C.M = 120

⇒ \(\frac{1}{R_{total}} = \frac{20\:+\:15\:+\:10}{120}\)

⇒ \(\frac{1}{R_{total}} = \frac{45}{120}\)

∴ \( \scriptsize R_{total} = \normalsize \frac{120}{45}\)

⇒ \( \scriptsize R_{total} = 2.67 \Omega\)

Example 2:

Calculate the total resistance of the circuit below.

resistors in parallel

Solution:

Data given in the question:

R1 = 2 Ω, R2 = 6 Ω, R3 = 3 Ω

Formula: \(\frac{1}{R_{total}} = \frac{1}{R_1} \: + \: \frac{1}{R_2} \: + \: \frac{1}{R_3}\)

Substitute the values into the equation:

⇒ \(\frac{1}{R_{total}} = \frac{1}{2}+ \frac{1}{6} + \frac{1}{3}\)

L.C.M = 12

⇒ \(\frac{1}{R_{total}} = \frac{6\:+\:2\:+\:4}{12}\)

⇒ \(\frac{1}{R_{total}} = \frac{12}{12}\)

⇒ \(\frac{1}{R_{total}} = \frac{1}{1}\)

∴ \( \scriptsize R_{total} = 1 \Omega\)

Example 3:

Find the total resistance in the given circuit below.

Resistors in series and parallel

Solution:

Rtotal = Total Resistance in series + Total Resistance in parallel

Total Resistance in series, Rseries = 2 Ω + 3 Ω = 5 Ω

Total Resistance in parallel ⇒ \(\frac{1}{R_{parallel}} = \frac{1}{5}\: +\: \frac{1}{5} \\ = \frac{1}{R_{parallel}} = \frac{2}{5} \\ \scriptsize R_{parallel} = \normalsize \frac{5}{2} \\ \scriptsize R_{parallel} = 2 \frac{1}{2} \Omega\)

The resultant resistive circuit now looks something like this:

circuit

We can see that the two remaining resistances are connected together in a “SERIES” combination and again they can be added together.

Rtotal = Rseries + Rparallel

= \( \scriptsize 5\: \Omega \: + \: 2 \frac{1}{2}\: \Omega \)

Rtotal = \( \scriptsize 7 \frac{1}{2}\: \Omega \)

Example 3:

Calculate the total current ( IT ) taken from the 12V supply in the following circuit.

resistance 1 1

Data given in the question:

R1 = 6 Ω, R2 = 8 Ω, R3 = 4 Ω, R4 = 12 Ω

Voltage = 12V, IT = ?

R2 and R3 are connected together in series so we can add them together to produce the aggregate resistance.

R2 + R3 = 8Ω + 4Ω = 12Ω

We can replace both resistor R2 and R3 above with a single resistor Rx of resistance value 12Ω

resistance in parallel and series

From this new arrangement, we can now see resistor Rx is parallel with the resistor R4.

We can calculate the total resistance of the parallel circuits using the formula;

Formula: \(\frac{1}{R_{total}} = \frac{1}{R_x} \: + \: \frac{1}{R_4} \)

⇒ \(\frac{1}{R_{total}} = \frac{1}{12} \: + \: \frac{1}{12} \)

⇒ \(\frac{1}{R_{total}} = \frac{2}{12} \)

⇒ \(\frac{1}{R_{total}} = \frac{1}{6} \)

∴ \( \scriptsize R_{total} = 6 \: \Omega \)

The resultant resistive circuit now looks something like this:

ex3

From the diagram the remaining tow resistors are connected in series, so we add them to get the total resistance of the circuit

Rtotal = R1 + Ry

Rtotal = 6Ω + 6Ω

Rtotal = 12Ω

By using Ohm’s Law, the value of the current ( I ) flowing around the circuit is calculated as:

I = \( \frac{V}{R}\)

V = 12V, R = 12Ω

I = \( \frac{12}{12}\)

I = 1 Ampere (A)

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