Question

3 resistors of 3 Ω, 6 Ω and 8 Ω are in series with a 2V battery. Across which resistor is the potential difference minimum?

Answer 1

$\bigstar\underline\mathtt{Concept }$

✏️Here the concept of Ohms Law is used. According to the ohm's law the current passing through a metallic element is directly proportional to the potential difference across its and provided the temperature remains constant.

✏️The total potential difference across the combination of resistor in series is equal to the sum of potential difference across the individual resistor.

${\underline{ \boxed {\mathtt{V = V_1 + V_2 + V_3....+V_n}}}}$

✏️The current across each resistor remains the same in series arrangement.

$\bigstar\underline\mathtt{Solution}$

The total resistance in series arrangement is given by :-

${\underline{\boxed{\tt{R_eq= R_1+R_2+R_3...+R_n}}}}$

Now we will calculate the total resistance

$\sf{\implies \: R_eq= 3+6+8} \\ \\ \sf{\implies R_eq= 17Ω}$

→Total voltage is given as 2 V

✿According to ohm's Law

${\underline{\boxed{\mathtt{V= IR}}}}$

where

• v denotes voltage
• I denotes current
• R denotes resistance

$\sf{ \implies I = \frac{2}{17}} \\ \\ \sf{ \implies I = 0.117 A}$

$\mathtt{ \bullet Now \: the \: potential \: difference \: across \: 3Ω \: resistor}$

$\sf{\to V= IR} \\ \\ \sf{ \to V= 0.117 × 3} \\ \\ \sf{ \to v= 0.351 v}$

$\mathtt{ \bullet Now \: the \: potential \: difference \: across \: 6Ω \: resistor}$

→V= IR

→V= 0.117× 6

→V= 0.702 v

$\mathtt{ \bullet Now \: the \: potential \: difference \: across \: 8Ω \: resistor}$

→V= IR

→V= 0.117 × 8

→v=0.936 v

______________________

$\mathtt{\therefore \: the \: resistor \: with \: lowest \: potential \: difference \: is \: 3Ω }$

✷Verification

${ \underline{ \sf{V_{eq}= V_1+V_2+V_3 }}}\\ \\ \sf{ \implies V_{eq}= 0.351+ 0.702+ 0.936 }\\ \\ \sf{ \implies V_{eq}= 1.989 \: v\approx \: 2v}$

----------------------------------------