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Resistance in parallel connection: Resistance in parallel connection: Steady State Solution of DC Circuits and Problems based on ohm’s law

**Steady State Solution of DC Circuits:**

**Resistance in series connection:**

The
resistors R_{1}, R_{2}, R_{3} are connected in series
across the supply voltage “V”. The total current flowing through the circuit is
denoted as “I”. The voltage across the resistor R_{1}, R_{2}
and R_{3} is V_{1}, V_{2}, and V_{3}
respectively.

V_{1}
= I*R_{1} (as per ohms law)

V_{2}=
I*R_{2}

V_{3}
= I*R_{3}

V = V_{1}+V_{2}+V_{3}

= IR_{1}+IR_{2}+IR_{3}

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

**R = R _{1}+R_{2}+R_{3}**

**Resistance in parallel connection:**

The
resistors R_{1}, R_{2}, R_{3} are connected in parallel
across the supply voltage “V”. The total current flowing through the circuit is
denoted as “I”. The current flowing through the resistor

R_{1},
R_{2} and R_{3} is I_{1}, I_{2}, and I_{3}
respectively.

I = V / R
(as per ohms law)

I _{1}
= V_{1} / R_{1}

I_{2}
= V_{2} / R_{2}

I_{3}
= V_{3} / R_{3}

V_{1}
= V_{2} = V_{3} = V

From the
above diagram

I = I_{1}+I_{2}+I_{3}

= V_{1}
/ R_{1} + V_{2} / R_{2} + V_{3} / R_{3}

= V / R_{1}+
V/R_{2} +V/R_{3}

I = V (1/R_{1} +1/R_{2} +1/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}**

__Problems based on ohm’s law__

**Problem 1:**

A current of 0.5 A is flowing through the resistance of 10Ω.Find the
potential difference between its ends.

__Given data:__

Current I= 0.5A.

Resistance
R=1Ω

__T o f i n d__

Potential
difference V = ?

__Formula used:__

V = IR

__Solution__**:**

V = 0.5 ×
10 = 5V.

**Result :**

The
potential difference between its ends = 5 V

**Problem :2**

A supply voltage of 220V is applied to a 100 Ω resistor. Find the
current flowing through

it.

__Given data__

Voltage V
= 220V

Resistance R = 100Ω

__To find:__

Current I
= ?

__Formula used:__

Current I
= V / R

__Solution:__

Current I
= 220/100

= 2.2 A

__Result:__

The
current flowing through the resistor = 2.2 A

**Problem : 3**

Calculate the resistance of the conductor if a current of 2A flows
through it when the potential difference across its ends is 6V.

__Given data__

Current I
= 2A

Voltage V
= 6V

__To find:__

Resistance
R = ?

__Formula used:__

Resistance
R = V / I

__Solution:__

Resistance
R = 6 / 2

= 3 Ω

__Result:__

The value
of resistance R = 3Ω

**Problem: 4**

Calculate the current and resistance of a 100 W, 200V electric bulb**.**

__Given data:__

Power P =
100W

Voltage V
= 200V

__To find:__

Current I
=?

Resistance
R =?

__Formula used:__

Power P =
V *I

Current I
= P / V

Resistance
R = V / I

__Solution:__

Current I
= P / V

= 100 / 200

= 0.5 A
Resistance R = V / I

= 200 / 0.2

= 400 Ω

__Result:__

The value
of the current I = 0.5 A

The value
of the Resistance R = 400 Ω

**Problem: 5**

A circuit is made of 0.4 Ω wire, a 150Ω bulb and a 120Ω rheostat
connected in series. Determine the total resistance of the circuit.

__Given data:__

Resistance
of the wire = 0.4Ω

Resistance of bulb =
1 5 0 Ω

Resistance of rheostat = 120Ω

__To find:__

The total
resistance of the circuit R _{T} =?

__Formula used:__

The total
resistance of the circuit R _{T} = R_{1}+R_{2}+R_{3}

_{Solution:}

_{Total
resistance ,R = 0.4 + 150 +120}

_{ = 270.4Ω}

_{Result:}

The total
resistance of the circuit R _{T} = 270.4 Ω

**Problem 6:**

Three resistances of values 2Ω, 3Ω and 5Ω are connected in series
across 20 V, D.C supply

.Calculate (a) equivalent resistance of the circuit (b) the total
current of the circuit (c) the voltage drop across each resistor and (d) the
power dissipated in each resistor.

__Given data:__

R_{1}
= 2Ω

R_{2}
= 3Ω

R_{3}
= 5Ω

V = 20V

__To find:__

R _{T}
=?

I _{T}
=?

V_{1},
V_{2}, V_{3} =?

P_{1},
P_{2}, P_{3} =?

**Formula used:**

R_{T}
= R_{1}+R_{2}+R_{3} (series connection)

I_{T}
= V_{T} / R_{T}

V_{1}
= R_{1}*I_{1}

V_{2}=
R_{2}*I_{2}

V_{3}
= R_{3}*I_{3}

P_{1}=V_{1}*I_{1}

P_{2}=V_{2}*I_{2}

P_{3}=V_{3}*I_{3}

**Solution:**

R_{T} = R_{1}+R_{2}+R_{3}
= 2+3+5

**R _{T} = 10Ω**

I_{T} = V_{T} / R_{T} = 20
/ 10

**I _{T} = 2 A**

In series
connection I_{1} = I_{2} = I_{3} = I_{T} = 2A

V_{1} = I_{1}*R_{1} = 2*2

**V _{1} = 4 V**

V_{2} = I_{2}*R_{2} = 2*3

**V _{2} = 6 V**

V_{3} = I_{3}*R_{3} = 5*2

**V _{3} = 10V**

P_{1}
= V_{1}*I_{1}

= 4*2

**P _{1} = 8W**

P_{2}
= V_{2}*I_{2}

= 6*2

**P _{2} = 12W**

P_{3} = V_{3}*I_{3} = 10*2

**P _{3} = 20W**

**Result:**

(a).
Equivalent resistance of the circuit R_{T} = 10Ω

(b). The
total current of the circuit I_{T}
= 2A

(c).
Voltage drop across each resistor V_{1} = 4 V, V_{2} = 6 V, V_{3}
= 10V

(d). The
power dissipated in each resistor P_{1} = 8W, P_{2} = 12W, P_{3}
= 20W

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