Solving Clipper Circuit containing battery and resistance in series
A simple way to approach this problem is to look at each half cycle of the input.

Consider the positive half cycle.

Only D1 will conduct so we can ignore the D2 branch of the circuit. D1 can only conduct if the input voltage is greater than B1 (5V) + the forward voltage drop across D1 (say 0V for a 'perfect diode', 0.2V for Germanium and 0.7V for Silicon).

The output voltage will simply follow the input voltage below this level (as there is no load on the output, i.e open circuit)

Above this 'clipping' level (B1 + Vdiode) some voltage will be dropped across R and some voltage will be dropped across R1. In this case R = R1 so these voltages will be equal.

Example calculations:

Suppose we have a 'perfect' diode and the input is +6V. 

R + R1 will have 6 - 5V (=1v) across them. You could use Ohm's law to calculate the current through (R + R1) but simple inspection would tell you as the resistors have equal value each one will drop half that voltage, 0.5V. So the output for a +6V input will be +5.5V. [or for Ge = 5.2 + 0.4V = 5.6V and for Si = 5.7 + 0.15V = 5.85V]

Suppose the peak input was + 20V.

The resistors will have 20 - 5 (=15V) across both so each resistor will drop 7.5V making the output voltage 5 + 7.5  (or 20 - 7.5) = 12.5V [ for Ge = 5.2 + 7.4 = 12.6V and for Si = 5.7 + 7.15 = 12.85V)

By symmetry in this circuit the positive and negative half cycles give symmetrical answers but you could analyze non symmetrical circuits by treating each half cycle separately.