Question

2 ohm ,3 ohm,5ohm,4ohm battery of 5 volt are in parallel find resistance and current​

Answer 1

Given Information :

• 2Ω, 3Ω , 5Ω and 4Ω resistors and a battery of 5V are connected in parallel combination.

To calculate :

• Effective resistance (R)

• Current (I)

Calculation :

$\underline{\small \sf {\maltese \; \; \; Finding \: effective \: resistance \: (R) : \; \; \; }}$

When the resistors are connected in parallel combination, effective resistance is given by the formula given below :-

$\boxed { \sf {\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}+ \dfrac{1}{R_3} + \dots \dfrac{1}{R_n}}}$

Here,

• $\sf {R_1 }$ = 2Ω
• $\sf {R_2 }$ = 3Ω
• $\sf {R_3 }$ = 5Ω
• $\sf {R_4 }$ = 4Ω

Substituting values,

$\longmapsto \sf{ \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}+ \dfrac{1}{R_3} + \dfrac{1}{R_4 } }$

$\longmapsto \sf{ \dfrac{1}{R} = \dfrac{1}{2} + \dfrac{1}{3}+ \dfrac{1}{5} + \dfrac{1}{4 } }$

$\longmapsto \sf{ \dfrac{1}{R} = \dfrac{30 + 20 + 12 + 15}{60} }$

$\longmapsto \sf{ \dfrac{1}{R} = \dfrac{77}{60} }$

$\longmapsto \boxed { \pmb {\rm \red{ R= \dfrac{60}{77} \: \Omega }} }$

Therefore, resistance is 60/77 Ω.

$\underline{\small \sf {\maltese \; \; \; Finding \: current \: (I) : \; \; \; }}$

We have,

• Potential difference (V) = 5 V
• Resistance (R) = $\sf \dfrac{60}{77}$ Ω

By applying Ohm's law to the whole circuit,

$\boxed{\sf {V = I R}}$

$\longmapsto \sf { 5= I \times \dfrac{60}{77}}$

$\longmapsto \sf { 5 \div \dfrac{60}{77}= I }$

$\longmapsto \sf { 5 \times \dfrac{77}{60}= I }$

$\longmapsto \boxed { \pmb {\rm \red{\dfrac{77}{12}\: A= I }} }$

Therefore, current is 77/12 Ampere.