Model Paper 1 — Question 9

(i) & (iii)

Reason —

(i) Here all resistances are in parallel combination and have same resistance value equal to 3 Ω.

Then,

$\dfrac{1}{R_{eq}}=\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{3}{3}=1 \\[1em] \Rightarrow R_{eq} = 1 \Omega$

(ii) In this case two resistances of value 3 Ω are in parallel combination and this combination is in series with 6 Ω resistance.

For parallel combination,

$R_p = \dfrac{3\times3}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\ Ω$

For series combination,

$R_{eq} = R_p+6=\dfrac{3}{2}+6=\dfrac{15}{2}\ Ω$

(iii) Here two 1 Ω resistances are in series combination and this whole combination is in parallel combination with itself.

For series combination,

$R_s=1+1=2\ Ω$

For Parallel combination,

$R_{eq} = \dfrac{2\times2}{2+2}=\dfrac{4}{4}= 1\ Ω$

(iv) Here all resistances are in series combination.

$R_{eq}=3+2+3=8\ Ω$