When a battery is connected to a \(100 .-\Omega\) resistor, the current is \(4.00 \mathrm{~A}\). When the same battery is connected to a \(400 .-\Omega\) resistor, the current is 1.01 A. Find the emf supplied by the battery and the internal resistance of the battery.

We are given the following information: - When connected to a \(100 \Omega\) resistor, the current is \(4.00 A\). - When connected to a \(400 \Omega\) resistor, the current is \(1.01 A\). Our goal is to find the emf supplied by the battery and the internal resistance of the battery, which we will denote by \(E\) and \(r\), respectively.

First, let's recall Ohm's law, which states that the voltage across a resistor is directly proportional to the current flowing through it: $$V = IR$$ Now, let's consider the equations for the internal resistance of the battery. When a battery is connected to a resistor, the total resistance in the circuit is \(R + r\) (the external resistor plus the internal resistance of the battery). Therefore, the current flowing through the circuit can be written as: $$I = \frac{E}{R + r}$$ Using the given values for the two cases, we can set up a system of equations as follows: Case 1: \(100 \Omega\) resistor and \(4.00 A\) current $$4.00 = \frac{E}{100 + r}$$ Case 2: \(400 \Omega\) resistor and \(1.01 A\) current $$1.01 = \frac{E}{400 + r}$$

Now, we have two equations with two unknowns, \(E\) and \(r\). We can solve this system of equations using the method of substitution or elimination. In this case, let's use the elimination method. First, we will multiply both equations by suitable factors to make their coefficients equal. Equation 1 multiplied by \(400+r\): $$(4.00)(400 + r) = E$$ Equation 2 multiplied by \(100+r\): $$(1.01)(100 + r) = E$$ Now, since both equations are equal to \(E\), we can set them equal to each other and solve for \(r\): $$(4.00)(400 + r) = (1.01)(100 + r)$$ Expanding and simplifying the equation: $$1600 + 4.00r = 101 + 1.01r$$ Subtract \(1.01r\) and \(101\) from both sides: $$3.99r = 1499$$ Divide by \(3.99\): $$r \approx 376.19 \Omega$$ Now substitute the value of \(r\) back into one of the original equations to find \(E\). Let's use the first equation: $$4.00 = \frac{E}{100 + 376.19}$$ Multiply both sides by \(100 + 376.19\): $$E \approx 1903.19 \mathrm{V}$$

From our calculations, the emf supplied by the battery is approximately \(1903.19 V\), and the internal resistance of the battery is approximately \(376.19 \Omega\).