Question

Calulate the equivalent resistance between points A and C.
12
С
312
223
310
А
B
222​

Answer 1

Answer:

Equivalent resistance between A and C is 1Ω

Step-by-step explanation:

In the given question, we are asked to find the equivalent resistance between points A and C. There are three wires between A and C( ADC, ABC and AC).

All the three wires are in parallel connection to each other.

We have formula for parallel connection ::

$\sf : {\implies \dfrac{1}{R_{(Equivalent)}} = \dfrac 1 R_1 + \dfrac{1}{R_2} + \dfrac{1}{R_3} \dots upto \: n}$

We have formula for series connection ::

$\sf {: \implies R_{(Equivalent)}= R_1 + R_2 + R_3 \dots \:upto\:n}$

Resistors between AD and DC is series connection.

$\sf {:\implies Resistance_{(ADC)} = R_1 + R_2}$

$\sf {:\implies Resistance_{(ADC)} = 1 \Omega + 2 \Omega}$

$\sf {:\implies Resistance_{(ADC)} = 3 \Omega}$

‎ ‎ ‎

Resistors between AB and BC is series connection.

$\sf {:\implies Resistance_{(ABC)} = R_1 + R_2}$

$\sf {:\implies Resistance_{(ABC)} = 1 \Omega + 2 \Omega}$

$\sf {:\implies Resistance_{(ABC)} = 3 \Omega}$

Wires ABC , ADC and AC are parallel connection.

$\sf : {\implies \dfrac{1}{R_{(Equivalent)}} = \dfrac 1 R_1 + \dfrac{1}{R_2} + \dfrac{1}{R_3}}$

$\sf : {\implies \dfrac{1}{R_{(Equivalent)}} = \dfrac 1 {3\Omega}+ \dfrac{1}{3\Omega} + \dfrac{1}{3\Omega}}$

$\sf : {\implies \dfrac{1}{R_{(Equivalent)}} = \dfrac{1+1+ 1 }{3\Omega}}$

$\sf : {\implies \dfrac{1}{R_{(Equivalent)}} = \dfrac{3}{3\Omega}}$

$\sf {:\implies R_{(Equivalent)} = 1 \Omega}$

Answer 2

Answer:

Answer:

Equivalent resistance between A and C is 1Ω

Step-by-step explanation:

In the given question, we are asked to find the equivalent resistance between points A and C. There are three wires between A and C( ADC, ABC and AC).

All the three wires are in parallel connection to each other.

We have formula for parallel connection ::

\sf : {\implies \dfrac{1}{R_{(Equivalent)}} = \dfrac 1 R_1 + \dfrac{1}{R_2} + \dfrac{1}{R_3} \dots upto \: n}:⟹

R

(Equivalent)

1

=

R

1

1

+

R

2

1

+

R

3

1

…upton

We have formula for series connection ::

\sf {: \implies R_{(Equivalent)}= R_1 + R_2 + R_3 \dots \:upto\:n}:⟹R

(Equivalent)

=R

1

+R

2

+R

3

…upton

Resistors between AD and DC is series connection.

\sf {:\implies Resistance_{(ADC)} = R_1 + R_2}:⟹Resistance

(ADC)

=R

1

+R

2

\sf {:\implies Resistance_{(ADC)} = 1 \text{\O}mega + 2 \text{\O}mega}:⟹Resistance

(ADC)

=1Ømega+2Ømega

\sf {:\implies Resistance_{(ADC)} = 3 \text{\O}mega}:⟹Resistance

(ADC)

=3Ømega

‎ ‎ ‎

Resistors between AB and BC is series connection.

\sf {:\implies Resistance_{(ABC)} = R_1 + R_2}:⟹Resistance

(ABC)

=R

1

+R

2

\sf {:\implies Resistance_{(ABC)} = 1 \text{\O}mega + 2 \text{\O}mega}:⟹Resistance

(ABC)

=1Ømega+2Ømega

\sf {:\implies Resistance_{(ABC)} = 3 \text{\O}mega}:⟹Resistance

(ABC)

=3Ømega

Wires ABC , ADC and AC are parallel connection.

\sf : {\implies \dfrac{1}{R_{(Equivalent)}} = \dfrac 1 R_1 + \dfrac{1}{R_2} + \dfrac{1}{R_3}}:⟹

R

(Equivalent)

1

=

R

1

1

+

R

2

1

+

R

3

1

\sf : {\implies \dfrac{1}{R_{(Equivalent)}} = \dfrac 1 {3\text{\O}mega}+ \dfrac{1}{3\text{\O}mega} + \dfrac{1}{3\text{\O}mega}}:⟹

R

(Equivalent)

1

=

3Ømega

1

+

3Ømega

1

+

3Ømega

1

\sf : {\implies \dfrac{1}{R_{(Equivalent)}} = \dfrac{1+1+ 1 }{3\text{\O}mega}}:⟹

R

(Equivalent)

1

=

3Ømega

1+1+1

\sf : {\implies \dfrac{1}{R_{(Equivalent)}} = \dfrac{3}{3\text{\O}mega}}:⟹

R

(Equivalent)

1

=

3Ømega

3

\sf {:\implies R_{(Equivalent)} = 1 \text{\O}mega}:⟹R

(Equivalent)

=1Ømega