24P3Q06

6a)

Similarity: Both refer to energy converted per unit charge, measured in volts.

Difference: Whie e.m.f. is about conversion from non-electrical to electrical forms by a battery, p.d. is about conversion from electrical to non-electrical forms in the external circuit.

6bi)

The voltmeter is not measuring the e.m.f. but the terminal p.d. of the battery.

Since a current is flowing, there is p.d. across the internal resistance of the battery, making the terminal p.d. smaller than the e.m.f.

6bii)

Total resistance\displaystyle \begin{aligned}   &=3.0+((6.0+4.0)//10) &=3.0+5.0 &=8.0\text{  }\Omega\!\!\text{ }\end{aligned}

P.d. across the 10 Ω \displaystyle \begin{aligned}&=(0.15)(8.0) &=1.2\text{ V}\end{aligned}

6ci)

6cii)

First, solve for the internal resistance of battery r.

\displaystyle \begin{aligned}{{V}_{t}}&=\varepsilon -Ir 1.2&=1.5-(0.15)r r&=2.0\text{  }\Omega\!\!\text{ }\end{aligned}

Effective resistance across the 10 Ω\displaystyle \begin{aligned}&=(6.0+9.0)//10 &={{(\frac{1}{15}+\frac{1}{10})}^{-1}} &=6.0\text{  }\Omega\!\!\text{ }\end{aligned}

By P.D.P., p.d. across the 10 Ω\displaystyle \begin{aligned}&=\frac{6.0}{6.0+3.0+2.0}\times 1.5 &=0.818\,\text{V}\end{aligned}

One thought on “24P3Q06”

  1. Pingback: A-Level 2024 – xmPhysics

Leave a comment