Question

A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp.

Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.

Answer 1

Let the resistance of the electric lamp be R_lamp.

Current (I) = 1 A

Resistance of conductor (R_conductor)  = 5Ω

Potential difference of battery (V) = 10 V  

Since the lamp and the conductor is connected in series, thus same current 1 A will pass through both of them.

Using Ohm's law, 

`"R"_"net" = "V"/"I"`

`"R"_"net" = 10/1`

`"R"_"net" = 10`Ω

`"R"_"net" = "R"_"lamp" + "R"_"conductor"`

⇒ 10 = R_lamp + 5

⇒ R_lamp = 5Ω

Potential difference across lamp,

V_lamp = I × R_lamp = 1 × 5 = 5V

When 10Ω resistor is connected parallel to the series combination of lamp and conductor (R_net = 5 + 5 = 10Ω) then the equivalent resistance,

`1/("R"_"eq") = 1/10 + 1/10 = 2/10 =1/5`

⇒ R_eq = 5Ω

Using Ohm's law

`"I"^"'" = "V"/"R"_"eq"`

⇒ `"I"^"'" = 10/5`

⇒ `"I"^"'" = 2  "A"`

Current will distribute equally in two parallel parts.

Thus, `"I"^"'"/2 = 1"A"` current will pass through both the lamp and the resistor of 5Ω (because they are connected in series).

Potential difference across the lamp (R_lamp) = 5Ω

`"V"_"lamp"^"'" = 1 xx 5 = 5"V"`

Hence, there will be no change in current through the conductor of resistance 5Ω, and potential difference across the lamp.