An electric circuit has two power sources. The voltage, V₁, provided by the first power source, at time t, is modelled by V₁ = Re(2e³ᵗⁱ) The voltage, V₂, provided by the second power source is modelled by V₂ = Re(5e⁽³⁺⁴⁾ⁱ) The total voltage in the circuit, V_T, is given by V_T = V₁ + V₂.

METHOD 1 recognizing that the real part is distributive $V_{T}=\operatorname{Re}\left(2 \mathrm{e}^{3 i \mathrm{t}}+5 \mathrm{e}^{3 i+4 \mathrm{i}}\right)$

$=\operatorname{Re}\left(\mathrm{e}^{3 i}\left(2+5 \mathrm{e}^{4 \mathrm{i}}\right)\right)$ (M1) (from the GDC) $2+5 \mathrm{e}^{4 \mathrm{i}}=3.99088 \ldots \mathrm{e}^{-1.89418 . . \mathrm{i}}$ (A1) Note: Accept arguments differing by $2 \pi$ e.g. $4.38900 \ldots$...). therefore $V_{T}=3.99 \cos (3 t-1.89) \quad (3.99088 \ldots \cos (3 t-1.89418 \ldots))$ (A1) Note: Award the last $\boldsymbol{A1}$ for the correct values of $A, B$ and $C$ seen either in the required form or not. If method used is unclear and answer is partially incorrect, assume Method 2 and award appropriate marks eg. (M1)A1A0AO if only $A$ value is correct.

METHOD 2 converting given expressions to cos form $V_{T}=2 \cos 3 t+5 \cos (3 t+4)$ (M1) (from graph) $A=3.99$ (3.99088...) (A1) $V_{T}=3.99 \cos (B t+C)$ either by considering transformations or inserting points $B=3$ (A1) $C=-1.89(-1.89418 \ldots)$ (A1) Note: Accept arguments differing by $2 \pi$ e.g. $4.38900 \ldots$ (so, $\left.V_{T}=3.99 \cos (3 t-1.89)(3.99088 \ldots \cos (3 t-1.89418 \ldots))\right)$ Note: It is possible to have $A=3.99, B=-3$ with $C=1.89$ OR $A=-3.99, B=3$ with $C=1.25$ OR $A=-3.99, B=-3$ with $C=-1.25$ due to properties of the cosine curve.

maximum voltage is $3.99(3.99088 \ldots$ ) (units) (A1)