+-2.5 points OSUniPhys1 27.3.WA.030.Tutorial. Find the potential difference between point a and point b for the situation shown below. Here $\mathcal{E}_1 = 12.0 \text{ V}$, $\mathcal{E}_2 = 8.27 \text{ V}$ and $R_1 = 4.13 \Omega$, $R_2 = 6.04 \Omega$, and $R_3 = 1.88 \Omega$. $V_a - V_b = \text{____} \text{ V}$
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Potential difference, also known as voltage, is the difference in electric potential energy per unit charge between two points in an electric circuit. In this situation, we are given the values of V_a and V_b, which represent the potential at points a and b, Show more…
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Find the potential difference between points- $P$ and $-Q$ Which point is at the higher potential? From the result of the currents through $P$ and $Q$ are $$ \begin{array}{l} I_{P}=\frac{2 \Omega+18 \Omega}{10 \Omega+5 \Omega+2 \Omega+18 \Omega}(7.0 \mathrm{~A})=4.0 \mathrm{~A} \\ I_{Q}=\frac{10 \Omega+5 \Omega}{10 \Omega+5 \Omega+2 \Omega+18 \Omega}(7.0 \mathrm{~A})=3.0 \mathrm{~A} \end{array} $$ Now we start at point- $P$ and go through point- $a$ to point- $Q$, to find Voltage change from $P$ to $Q=+(4.0 \mathrm{~A})(10 \Omega)-(3.0 \mathrm{~A})(2 \Omega)=+34 \mathrm{~V}$ (Notice that we go through a potential rise from $P$ to $a$ because we are going against the current. From $a$ to $Q$ there is a drop.) Therefore, the voltage difference between $P$ and $Q$ is $34 \mathrm{~V}$, with $Q$ being at the higher potential.
(a) Find the potential of point a with respect to point $b$ in Fig. $\mathrm{P} 26.67$ . (b) If points $a$ and $b$ are connected by a wire with negligible resistance, find the current in the $12.0-\mathrm{V}$ battery.
In the figure, R1 = 100 Ω, R2 = 50 Ω, and the ideal batteries have emfs E1 = 6.0 V, E2 = 5.0 V, and E3 = 4.0 V. Find the current in resistor 1, the current in resistor 2, and the potential difference between points a and b.
Madhur L.
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