1. Problem 1: A 3V battery with internal resistance $r=0.5\ \Omega$ is connected to a load resistance $R=4\ \Omega$. Find: a) Current delivered $I$ b) Terminal potential difference $V_T$ c) Power developed in emf $P_{emf}$ d) Power lost in internal resistance $P_r$ e) Power lost in external load $P_R$ 2. Problem 2: A cell of emf $E=2V$ and internal resistance $r=1\ \Omega$ connected to a resistor $R=9\ \Omega$. Find the same quantities as above. --- **Step 1: Formula and rules** The current delivered by the battery is given by Ohm's law for the entire circuit: $$I = \frac{E}{R + r}$$ Terminal potential difference $V_T$ is the voltage across the external load: $$V_T = IR$$ Power developed by emf: $$P_{emf} = EI$$ Power lost in internal resistance: $$P_r = I^2 r$$ Power lost in external load: $$P_R = I^2 R$$ --- ### Problem 1 1. Calculate current $I$: $$I = \frac{3}{4 + 0.5} = \frac{3}{4.5} = \frac{\cancel{3}}{\cancel{4.5}} = 0.6667\ A$$ 2. Terminal potential difference $V_T$: $$V_T = I R = 0.6667 \times 4 = 2.6667\ V$$ 3. Power developed in emf: $$P_{emf} = E I = 3 \times 0.6667 = 2.0001\ W$$ 4. Power lost in internal resistance: $$P_r = I^2 r = (0.6667)^2 \times 0.5 = 0.4444 \times 0.5 = 0.2222\ W$$ 5. Power lost in external load: $$P_R = I^2 R = (0.6667)^2 \times 4 = 0.4444 \times 4 = 1.7776\ W$$ --- ### Problem 2 1. Calculate current $I$: $$I = \frac{2}{9 + 1} = \frac{2}{10} = 0.2\ A$$ 2. Terminal potential difference $V_T$: $$V_T = I R = 0.2 \times 9 = 1.8\ V$$ 3. Power developed in emf: $$P_{emf} = E I = 2 \times 0.2 = 0.4\ W$$ 4. Power lost in internal resistance: $$P_r = I^2 r = (0.2)^2 \times 1 = 0.04\ W$$ 5. Power lost in external load: $$P_R = I^2 R = (0.2)^2 \times 9 = 0.04 \times 9 = 0.36\ W$$ --- **Summary:** Problem 1: $I=0.6667\ A$, $V_T=2.6667\ V$, $P_{emf}=2.0001\ W$, $P_r=0.2222\ W$, $P_R=1.7776\ W$ Problem 2: $I=0.2\ A$, $V_T=1.8\ V$, $P_{emf}=0.4\ W$, $P_r=0.04\ W$, $P_R=0.36\ W$