Question

Two resistors of values 50 ohm and 100ohm are connected in parallel. A 3rd resistor of
60 ohm is connected in series across the combination. Find the total resistance of the combination

Answer 1

Answer:

280/3 ohm

Explanation:

for parallel combination net resistance is given by formula R1R2/R1+R2

thus net resistance for parallel combination is 100/3 ohm

for series u know net resistance is R1+R2

thus

for whole combination net resistance is 100/3+60 which is equal to 280/3 ohms

Answer 2

Answer:

$\bigstar{\bold{Total\:resistance=93.33\:\Omega}}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Resistors R₁ ( 50 Ω) and R₂ (100 Ω) are connected inn parallel
• Resistor R₃ ( 60 Ω) is connected with the above combination

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Total resistance of the combination

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First we have to find the total resistance of the parallel combination

→ Resistance between resistors  connected in parallel is given by

  $\dfrac{1}{R} =\dfrac{1}{R_1} +\dfrac{1}{R_2}$

→ Substituting the datas we get,

  $\dfrac{1}{R} =\dfrac{1}{50} +\dfrac{1}{100}$

→ Simplifying we get,

  $\dfrac{1}{R}=\dfrac{2+1}{100}$

  $\dfrac{1}{R}=\dfrac{3}{100}$

  R = 100/3

  R = 33.33 Ω = R₄

→ Now we have to find the resistance across R₃ and R₄ connected in series.

→ The effective resistance across resistors in series is given by

   R = R₃ + R₄

→ Substituting the datas we get,

   R = 33.33 + 60

   R = 93.33 Ω

→ Hence the total resistance is 93.33 Ω

$\boxed{\bold{Total\:resistance=93.33\:\Omega}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The effective resistance when resistors are connected in series is given by

$R=R_1+R_2+R_3+........+R_n$

→ The effective resistance when resistors are connected in parallel is is given by

$\dfrac{1}{R} =\dfrac{1}{R_1} +\dfrac{1}{R_2}+\dfrac{1}{R_3}+....\dfrac{1}{R_n}$