**Resistors in series:**

A circuit is said to be a series circuit if the current flowing through all the elements is the same.

In figure three resistors R

_{1}, R_{2}, R_{3}are connected in series. So the current (I ) flowing through each resistor is same. But the voltage drop across each resistor is different due to its different resistance value. The total voltage in this circuit is ‘V’
According to Ohm’s Law , V = IR

Voltage drop across R

_{1}is V_{1}= IR_{1}
Voltage drop across R

_{2}is V_{2}= IR_{2}
Voltage drop across R

_{3}is V_{3}= IR_{3}
Total voltage

V = V

_{1}+ V_{2}+ V_{3}
= IR

_{1}+ IR_{2}+ IR_{3}
= I ( R

_{1}+R_{2}+ R_{3 })
IR = I ( R

_{1}+ R_{2 }+ R_{3})
R = ( R

_{1}+ R_{2}+ R_{3})
The total (or) equivalent resistance is equal to the sum of all the resistances which are connected in series

**Resistors in Parallel:**

A circuit is said to be a parallel circuit if the potential difference ( voltage ) across all the elements is the same.

In figure three resistors R1, R2, R3 are connected in parallel . So the potential drop across each resistor is same. But the current flowing through each resistor is different due its different resistance value. The total current I in the circuit divides to pass through the individual resistances.

According to Ohm’s Law

I = V/R

The current flowing through resistor R

_{1}is I_{1}= V/R_{1}
The current flowing through resistor R

_{2}is I_{2}= V/R_{2}
The current flowing through resistor R

_{3}is I_{3}= V/R_{3}
Total current I = I

_{1}+ I_{2}+ I_{3}= V/R_{1}+ V/R_{2}+ V/R_{3}
V/R = V(1/R

_{1}+ 1/R_{2}+ 1/R_{3})
1/R = 1/R

_{1}+ 1/R_{2}+ 1/R_{3}
The reciprocal of equivalent resistance is equal to the sum of resistances which are connected in parallel.

When two resistor R

_{1}and R_{2}are connected , then equivalent resistance is
R = R

_{1}R_{2}/ (R_{1}+ R_{2})**Resistors in Series - Parallel:**

A series parallel circuit is shown in figure. Here the resistors R

_{2}and R_{3}are connected in parallel. But the resistor R_{1}is connected in series with this parallel combination.
Equivalent resistance of the resistors R

_{2}andR_{3 }(parallel combination) is
1/ R

_{p}= 1/R_{2}+ 1/R_{3}
R

_{p}= R_{2 }R_{3}/ (R_{2}+ R_{3})
V = V

_{1}+ V_{2}
= IR

_{1}+ IR_{P}
= I ( R

_{1 }+R_{P})
IR = I ( R

_{1}+ R_{P})
R = R

_{1}+ R_{P}
R = R

_{1}+ R_{2}R_{3}/(R_{2}+ R_{3})
The current flowing through resistor R

_{2}is I_{2}= IR_{3}/ (R_{2}+ R_{3})
The current flowing through resistor R

_{3}is I_{3}= IR_{2}/(R_{2}+ R_{3})**Problem 1:**

In the circuit shown in figure , Find the current , the voltage across each resistor and the power dissipated in each resistor.

**Solution:**

Total resistance, R = R

_{1}+ R_{2}+ R_{3}= 6 + 9 + 10 = 25 Ω
I = V/R = 50/25 = 2 amps

Voltage drop across 6 Ω resistor, V

_{1}= IR_{1}= 2 x 6 = 12 volts
Voltage drop across 9 Ω resistor, V

_{2}= IR_{2}= 2 x 9 = 18 volts
Voltage drop across 10 Ω resistor, V

_{3}= IR_{3}= 2 x 10 = 20 volts
Power dissipated in 6 Ω resistor, P

_{1}= I^{2}R_{1}= 4 x 6 = 24 watts
Power dissipated in 9 Ω resistor, P

_{2}= I^{2}R_{2}= 4 x 9 = 36 watts
Power dissipated in 10 Ω resistor, P

_{3}= I^{2}R_{3}= 4 x 10 = 40 watts**Problem 2:**

Two resistors resistance 6 Ω and 12 Ω are connected in series. Find the equivalent Also find the equivalent resistance when they are connected in parallel.

Given data

Case 1: Series combination of 6 Ω and 12 Ω resistors.

R

_{1}= 6 Ω, R_{2}= 12 Ω**Solution:**

Equivalent resistance, R = R

_{1 }+ R_{2 }= 18 Ω
Case 2: Parallel combination of 6 Ω and 12 Ω resistors.

Equivalent resistance, 1/R = 1/ R

_{1 }+ 1/R_{2}
R = R

_{1}R_{2}/(R_{1}+ R_{2}) = 6 x 12/(6 + 12) = 72/18 = 4 Ω**Problem 3:**

In the circuit shown in figure , find the total resistance and current flowing through each branch.

Given data, R

_{1}= 5 Ω, R_{2}= 10 Ω, R_{3}= 20 Ω**Solution:**

Equivalent resistance, 1/R = 1/ R

_{1 }+ 1/R_{2}+ 1/R_{3 }= 0.2 + 0.1 + 0.05 = 0.35
Total resistance, R = 2.86 Ω

The current flowing through 5 Ω resistor, I

_{1}= V/R_{1}= 25/5 = 5 amps.
The current flowing through 10 Ω resistor, I

_{2}= V/R_{2}= 25/10 = 2.5 amps
The current flowing through 20 Ω resistor, I

_{3}= V/R_{3}= 25/20 = 1.25 amps**Problem 4:**

Resistors of values 2, 3, 4, and 5 ohm are connected in parallel. If the total power absorbed by all the resistors is 200w, find the voltage applied to the circuit.

Power, P = 200 watts

Power, P = 200 watts

R

_{1}= 2 Ω, R_{2}= 3 Ω, R_{3}= 4 Ω, R_{4}= 5 Ω**Solution:**

Total Resistance, 1/R = 1/ R

_{1 }+ 1/R_{2}+ 1/R_{3}+ 1/R_{4}= 0.5 + 0.33 +0.25 + 0.2
1/R = 1.28, R = 1/1.28 = 0.78 ohm

Power, P = V2/R

V

^{2}= PX R
V = √(P X R) = √ (200 X 0.78) = 12.5 volts

**Problem 5:**

Two resistors are connected in parallel across a 200V mains take a total current of 10amps. The power dissipated in one of the resistors is 1200. watts. Find the value of each resistor.

Voltage, V = 200 volts

Current, I = 10 amps

Power dissipated in R

_{1}of the resistors P_{1}= 1200 watts**Solution:**

I

_{1}= P_{1}/V = 1200/ 200 = 6 amps.
Resistors, R

_{1}= V/ I_{1}= 200/6 = 33.33 Ω
Total current, I = I

_{1 }+ I_{2}
10 = 6 + I

_{2}
I

_{2}= 4 amps
Resistors, R

_{2}= V/ I_{2}= 200/4 = 50 Ω
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