9.1 State Ohm's law in words - NSC Physical Sciences - Question 9 - 2018 - Paper 1

To find the reading on voltmeter V1, we use Ohm's law:

$V_1 = I A_2 imes R_{6 ext{Ω}}$ Substituting the values:

$V_1 = (0.6 ext{ A}) imes (4 ext{ Ω}) = 2.4 ext{ V}$

Using the total current through the circuit, we find V2:

$V_2 = I_{total} imes R_{5.8 ext{Ω}}$ Where total current $I_{total} = 0.4 ext{ A}$ (calculated earlier).

Thus,

$V_2 = (0.4 ext{ A}) imes (5.8 ext{ Ω}) = 2.32 ext{ V}$

The emf of the battery can be calculated considering the internal resistance:

$ext{Emf} = V_{ext} + I_{total} imes r$

Where $V_{ext} = (5.8 + 2.4) ext{ V} = 8.2 ext{ V}$. Thus:

$ext{Emf} = 8.2 ext{ V} + (0.6 ext{ A} imes 0.8 ext{ Ω}) = 9 ext{ V}$

Energy can be calculated using the formula:

$W = I^2 R t$ Where $I = 0.6 ext{ A}$ and $R = 0.8 ext{Ω}$. Thus:

$W = (0.6 ext{ A})^2 imes 0.8 ext{ Ω} imes 15 s = 5.76 ext{ J}$