Question

In the figure shown R1 =10 ohm, R2 = 30 ohms, R3=10 ohm, R4=20 ohm,r5=60ohm &a12v battery is connected to the arrangement. Calculate (a)the total resistance in the circuit, and (b) the total current flowing in the circuit. ​

Answer 1

Given :

In a circuit,

• R₁ = 10Ω
• R₂ = 30Ω
• R₃ = 10Ω
• R₄ = 20Ω
• R₅ = 60Ω

potential difference, v = 12 volts

To find :

(a) The total resistance in the circuit.

(b) The total current flowing in the circuit.

Solution :

(a) According to the question,

R₁ and R₂ are connected in parallel,

» The reciprocal of the combined resistance of a number of resistance connected in parallel is equal to the sum of the reciprocal of all the individual resistances. .i.e.,

$\leadsto \sf{\dfrac{1}{R_{12}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} }$

$\leadsto \sf{\dfrac{1}{R_{12}} = \dfrac{1}{10} + \dfrac{1}{30} }$

$\leadsto \sf{\dfrac{1}{R_{12}} = \dfrac{3 + 1}{30} }$

$\leadsto \sf{\dfrac{1}{R_{12}} = \dfrac{4}{30} }$

$\leadsto \sf{R_{12} = \dfrac{30}{4} }$

$\leadsto \sf{R_{12} = 7.5 \: ohms}$

Similarly, R₃, R₄ and R₅ are also connected in parallel,

$\leadsto \sf{\dfrac{1}{R_{345}} = \dfrac{1}{R_3} + \dfrac{1}{R_4} + \dfrac{1}{R_5} }$

$\leadsto \sf{\dfrac{1}{R_{345}} = \dfrac{1}{10} + \dfrac{1}{20} + \dfrac{1}{60} }$

$\leadsto \sf{\dfrac{1}{R_{345}} = \dfrac{6 + 3 + 1}{60} }$

$\leadsto \sf{\dfrac{1}{R_{345}} = \dfrac{10}{60} }$

$\leadsto \sf{R_{345} = \dfrac{60}{10} }$

$\leadsto \sf{R_{345} = 6 \: ohms}$

Now, R₁₂ and R₃₄₅ are connected in series,

» The combined resistance of any number of resistance connected in series is equal to the sum of the individual resistances. i.e.,

$\leadsto \sf R_{12345} = R_{12} + R_{345}$

$\leadsto \sf R_{eq} = 7.5 + 6$

$\leadsto \sf R_{eq} = 13.6 \: ohms$

thus, the total resistance of the circuit is 14.6 Ω.

(b) To find the current we can use ohm's law that is,

» At constant temperature the current flowing through a conductor is directly proportional to the potential difference across its ends that is,

V = RI

where,

• V denotes potential difference
• R denotes resistance
• I denotes current

substituting all the given values in the formula,

$\leadsto \sf V = RI$

$\leadsto \sf 12 = (13.6)I$

$\leadsto \sf I = \dfrac{12}{13.6}$

$\leadsto \sf I = 0.9 \: A$

thus, the total current flowing through the circuit is 0.9A.

Remember !

SI unit of resistance is ohms (Ω).

SI unit of current is ampere (A).