What is the value of resistor RR as we discussed earlier?


Answered 2 years ago
Answered 2 years ago
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We can start solution with Ohm's law for 10 Ω\Omega resistor. In parallel connection the potential difference across each element is the same:

I1=ΔVR1ΔV=R1I1ΔV=100.08ΔV=0.8 VI2=ΔVR2I2=0.815I2=0.053 AIin=IoutInet=I1+I2+I3I3=InetI1I2I3=20.080.053I3=1.867 AI3=ΔVRR=ΔVI3R=0.81.867R=0.428 Ω\begin{align*} I_1&=\frac{\Delta V}{R_1} \\ \Delta V&=R_1 I_1 \tag{multiply with $R_1$} \\ \Delta V&=10 \cdot 0.08 \tag{substitute} \\ \Delta V&=0.8 \text{ V} \\ I_2&=\frac{\Delta V}{R_2} \tag{Ohm's law for 15 $\Omega$ resistor} \\ I_2&=\frac{0.8}{15} \tag{substitute} \\ I_2&=0.053 \text{ A} \\ \sum I_{in}&=\sum I_{out} \tag{Kirchhoff's junction law} \\ I_{net}&=I_1+I_2+I_3 \tag{Kirchhoff's junction law for junction where 2.0 A enters} \\ I_3&=I_{net}-I_1-I_2 \tag{express $I_3$} \\ I_3&=2-0.08-0.053 \tag{substitute} \\ I_3&=1.867 \text{ A} \\ I_3&=\frac{\Delta V}{R} \tag{Ohm's law for unknown resistor} \\ R&=\frac{\Delta V}{I_3} \tag{express $R$} \\ R&=\frac{0.8}{1.867} \tag{substitute} \\ R&=\boxed{0.428 \text{ $\Omega$}} \end{align*}

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