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Hey you're getting wrong, the equivalent emf(voltage) of system will be 16V-8V because both have opposite poles facing each other, so their will be net flor of current (-ve to +ve) according to the cell of greater emf(16V cell).Then your total resistance is $$5+1.6+1.4 = 8 \Omega$$(all are in series) . $$I(Current) = E(e.m.f or Voltage)/R(Resistance) = 8/8 = ... 1 Let the resistance of the original wire be R. R = ρ (L/A) Now l = L/5 R’ = ρ (L/5A) Or R’ = R/5 Now 5 resistors of R’ are connected in parallel 1/R(net) = 5/R + 5/R + 5/R + 5/R + 5/R or 1/2 = 25/R or R = 50 Ω. -2 The wires have resistance. You had 5 resistors in series. If connected in parallel, they give a known parallel resistance. 1 Resistor is anything that pose resistance to flow of charges in a circuit. Here is how it looks like:$$R=\rho \dfrac lA Where $\rho$ is material property, $l$ is length and $A$ is cross sectional area. Here is how 5 resistors in parallel look like: and the circuit diagram showing the same: In parallel configuration voltage drop across each ...

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Seems like you're not making the connection between the actual physical setup and the equations. So, here's a translation: 1) A length of wire is a resistor. By resistor what we mean is that when we apply a potential difference $V$ between the two ends (like from a battery) the resulting current $I$ is given by $V = IR$ where $R$ is the resistance. 2) The ...

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$\Gamma_{ii}$ is the $i$-th entry on the diagonal of $\Gamma = L^+ = (D-A)^+$, $\Gamma_{jj}$ is the $j$-th entry, and $\Gamma_{ij}$ is the entry located at row $i$, column $j$. Thus $\Omega_{ij}$ is a scalar, but you could assemble all such values into a matrix $\Omega$ that gives the resistances between all pairs of vertices.

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