Energy Accounting (3-22-2020)

Electricity and Magnetism Level 3

Two AC voltage sources are connected across a resistor as shown. Let E 1 ( t ) E_1(t) be the cumulative energy supplied (outputted) by V 1 V_1 from time t = 0 t = 0 to time t t . E 2 ( t ) E_2(t) is the same, but for V 2 V_2 . Let E R ( t ) E_R(t) be the cumulative energy dissipated in the resistor from time t = 0 t = 0 to time t t .

Determine the value of the following quantity at time t = 1 t = 1 :

Q = E 1 E 2 ( E 1 + E 2 E R ) Q = \frac{E_1}{E_2 } \Big( \frac{E_1 + E_2 }{E_R } \Big)

Details and Assumptions:
1) V 1 ( t ) = sin ⁡ ( 2 t ) V_1(t) = \sin(2 t)
2) V 2 ( t ) = sin ⁡ ( t ) V_2(t) = \sin(t)
3) R = 1 R = 1
4) Be mindful of signs (positive and negative)


The answer is -1.585.

Steven Chase by Steven Chase , 37, USA

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1 solution

After some circuit analysis work, we get

E 1 ( t ) = 1 2 R ( t − 1 4 sin ⁡ 4 t − sin ⁡ t + 1 3 sin ⁡ 3 t ) ⟹ E 1 ( 1 ) ≈ 0.1973848208529 E_1(t)=\dfrac{1}{2R}\left (t-\dfrac{1}{4}\sin 4t-\sin t+\dfrac{1}{3}\sin 3t\right )\implies E_1(1)\approx 0.1973848208529

E 2 ( t ) = 1 2 R ( t − sin ⁡ t + 1 3 sin ⁡ 3 t − 1 2 sin ⁡ 2 t ) ⟹ E 2 ( 1 ) ≈ − 0.124539847767 E_2(t)=\dfrac{1}{2R}\left (t-\sin t+\dfrac{1}{3}\sin 3t-\dfrac{1}{2}\sin 2t\right )\implies E_2(1)\approx -0.124539847767

E R ( t ) = 1 2 R ( 2 t − 1 4 sin ⁡ 4 t − 2 sin ⁡ t + 2 3 sin ⁡ 3 t − 1 2 sin ⁡ 2 t ) ⟹ E R ( 1 ) ≈ 0.0728449730858 E_R(t)=\dfrac{1}{2R}\left (2t-\dfrac {1}{4}\sin 4t-2\sin t+\dfrac {2}{3}\sin 3t-\dfrac {1}{2}\sin 2t\right )\implies E_R(1)\approx 0.0728449730858 .

Therefore the given ratio is ≈ − 1.5849 \approx \boxed {-1.5849} .

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