Question

An electric lamp of resistance 10 ohm and a conductor of resistance 4 ohm are connected to a 6 volt battery as shown calculate the following
a- the total resistance of the circuit
b- the current through the circuit
c- the potential difference across the (i) electric lamp (ii) conductor
d- power of lamp​

Answer 1

Answer:

Given that,

Resistance of electric lamp (R₁) is 10 Ω.

Resistance of conductor (R₂) is 4 Ω.

Voltage of battery (V) is 6 V.

(a) The electric lamp and conductor are connected in series.

Total resistance of circuit = R₁ + R₂

= 10 Ω + 4 Ω

= 14 Ω

Thus,

Total resistance of circuit is 14 Ω.

(b) We know,

Ohm's law →

• I = V/R

[Where, I is current, V is Voltage and R is total resistance of circuit]

→ I = 6/14

→ I ≈ 0.42

Thus,

Current is 0.42 A.

(c) We know,

Potential difference :

V = IR

For Electric lamp :

→ V = 0.42 A × 10Ω

→ V = 4.2

Potential difference across electric lamp is 4.2 V.

For Conductor :

→ V = 0.14 A × 4 Ω

→ V = 1.68

Potential difference across conductor is 1.68 V.

(d) We know,

Electric power = I²R

→ Power of lamp = (0.42)² × 10 Ω

→ Power of lamp = 0.1764 × 10 Ω

→ Power of lamp = 1.764

Thus,

Power of lamp is 1.764 W.

Answer 2

Given that , An electric lamp has resistance of 10 Ω and a conductor of resistance of 4 Ω are connected to a 6 V ( or volt ) battery .

Need To Find : (i) The total resistance of the circuit , (ii) The current through the circuit , ( iii ) the potential difference across the (a) electric lamp (b) conductor & (iv) Power of lamp .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀¤ Finding (i) The Total Resistance of the Circuit :

As , We know that ,

⠀⠀⠀⠀▪︎⠀Formula for Total Resistance of Circuit :

$\qquad \star \:\:\underline {\boxed {\pink{\pmb{\frak{ \:\:Total \:Resistance \:\:(\:R_{eq}\:) \:=\:R_1 \:\: + \:R_2 \:\:}}}}}$

⠀⠀⠀⠀Here , $\sf R_1 \:\& \: R_2 \:$ are the Resistance of Electrical lamp and Conductor, respectively.

$\qquad \dashrightarrow \sf R_{eq} = \:\:R_1 \:\: + \:R_2\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf R_{eq} = \:\:R_1 \:\: + \:R_2\:\\\\\qquad \dashrightarrow \sf R_{eq} = \:\:4 \:\Omega\: + \:10 \:\Omega\:\\\\\qquad \dashrightarrow \sf R_{eq} = \:\:14 \:\Omega\:\\\\\qquad \dashrightarrow \underline {\boxed{\pmb{\frak { \purple{ \:\:Total \:Resistance \:\:(\:R_{eq}\:) = \:\:14 \:\Omega\:}}}}}$

$\qquad \therefore \:\underline {\sf Hence, \:Total \:Resistance \:in \:circuit \:\:is \:\pmb{\bf 14 \Omega \:}\:.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀¤ Finding (ii) The current through the circuit :

As , We know that ,

⠀⠀⠀⠀▪︎⠀Formula for Cuurrent ( I ) flowing through Circuit :

$\qquad \star \:\:\underline {\boxed {\pink{\pmb{\frak{ \:\:Current \:\:(\:I\:) \:=\:\dfrac{V}{R} \:\:}}}}}$

⠀⠀⠀⠀Here , V is the Voltage of Battery & R is the Total Resistance of Circuit.

$\qquad \dashrightarrow \sf I = \:\:\dfrac{V}{R}\:\\\\\qquad \dashrightarrow \sf I = \:\:\dfrac{6}{14}\:\\\\\qquad \dashrightarrow \sf I = \:\:0.42\:\\\\\qquad \dashrightarrow \underline {\boxed{\pmb{\frak { \purple{ \:\:Current \:\:(\:I\:) = \:\:0.42 \:A\:}}}}}$

$\qquad \therefore \:\underline {\sf Hence, \:Current \:flowing \:through\: \:circuit \:\:is \:\pmb{\bf 0.42 \:}\:.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀¤ Finding (iii) The potential difference across the (a) electric lamp (b) conductor :

As , We know that ,

⠀⠀⠀⠀▪︎⠀Formula for Potential Difference ( V ) flowing through Circuit :

$\qquad \star \:\:\underline {\boxed {\pink{\pmb{\frak{ \:\:Potential \:Difference \:\:(\:V\:) \:=\:IR \:\:}}}}}$

⠀⠀⠀⠀Here , I is the amount of Current & R is the Resistance of Conductor or Lamp .

• Potential Difference ( V ) of Lamp and Resistance of Lamp is 10 Ω

$\qquad \dashrightarrow \sf V = \:\:IR\:\\\\\qquad \dashrightarrow \sf V = \:\:0.42 \:\times 10 \:\\\\\qquad \dashrightarrow \sf V = \:\:4.2 \:\\\\\qquad \dashrightarrow \underline {\boxed{\pmb{\frak { \purple{ \:\:Potential \:Difference \:\:(\:V\:)_{(Lamp)} = \:\:4.2\:V\:}}}}}$

$\qquad \therefore \:\underline {\sf Hence, \:Potential \:Difference \:of \:Lamp\:is\: \: \:\pmb{\bf 4.2 V \:}\:.}$

• Potential Difference ( V ) of Conductor and Resistance of Conductor is 4 Ω

$\qquad \dashrightarrow \sf V = \:\:IR\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf V = \:\:IR\:\\\\\qquad \dashrightarrow \sf V = \:\:0.42 \:\times 4 \:\\\\\qquad \dashrightarrow \sf V = \:\:1.68 \:\\\\\qquad \dashrightarrow \underline {\boxed{\pmb{\frak { \purple{ \:\:Potential \:Difference \:\:(\:V\:)_{(Conductor)} = \:\:1.68\:V\:}}}}}$

$\qquad \therefore \:\underline {\sf Hence, \:Potential \:Difference \:of \:Conductor \:is\: \: \:\pmb{\bf 1.68 V \:}\:.}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀¤ Finding (iv) The Power of Lamp :

$\qquad \star \:\:\underline {\boxed {\pink{\pmb{\frak{ \:\:Power \:=\:I^2R \:\:}}}}}$

⠀⠀⠀⠀Here , I is the amount of Current & R is the Resistance Lamp .

$\qquad \dashrightarrow \sf Power_{(Lamp)} = \:\:I^2R\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\\\qquad \dashrightarrow \sf Power_{(Lamp)} = \:\:I^2R\:\\\\\qquad \dashrightarrow \sf Power_{(Lamp)} = \:\:(0.42)^2 \times 10 \:\\\\\qquad \dashrightarrow \sf Power_{(Lamp)} = \:\:1.764 \:\\\\\qquad \dashrightarrow \underline {\boxed{\pmb{\frak { \purple{ \:\:Power_{(Lamp)} = \:\:1.764\:W\:}}}}}$

$\qquad \therefore \:\underline {\sf Hence, \:Power \:of \:Lamp \:is\: \: \:\pmb{\bf 1.764 \:W \:}\:.}$