Question

Please solve this question with method

Answer 1

A 12A current flows through a resistor when 230V supply is connected to it.

resistance of resistor, R1 = (230/12) Ω = 19.167 Ω

Let R2 is the resistance is connected in parallel combination.

so, 1/Req = 1/R + 1/19.167Ω

Req = 19.167R/(R + 19.167)

now, Req = 230/16 = 14.375 Ω

⇒ 14.375 = 19.167R/(R + 19.167)

⇒14.375R + 14.375 × 19.167 = 19.167R

⇒14.135 × 19.167 = (19.167 - 14.375)R

⇒ R = 14.375 × 19.167/4.792

= 57.49 ≈ 57.5 Ω

hence, resistance of unknown resistor joined in parallel to increase current to 16A is 57.5Ω

Answer 2

$\huge\bold\purple{Answer:-}$

SP=Rs. 600

Loss=Rs. 100

Therefore,

1》CP=600+100

=Rs. 700

2》Loss%=Loss/CP×100

=100/700×100

=14.286%

HOPE IT HELPS❤❤